4.6 Graph Quadratic Functions Using Properties

Domenic Spilotro, MSc

4.6 Graph Quadratic Functions Using Properties

Learning Objectives

By the end of this section, you will be able to:

Recognize the graph of a quadratic function
Find the axis of symmetry and vertex of a parabola
Find the intercepts of a parabola
Graph quadratic functions using properties
Solve maximum and minimum applications

Try It

Before you get started, take this readiness quiz:

1) Graph the function $f (x) = x^{2}$ by plotting points.
2) Solve: $2 x^{2} + 3 x - 2 = 0$ .
3) Evaluate $- \frac{b}{2 a}$ when $a = 3$ and $b = - 6$ .

Recognize the Graph of a Quadratic Function

Previously we very briefly looked at the function $f (x) = x^{2}$ , which we called the square function. It was one of the first non-linear functions we looked at. Now we will graph functions of the form $f (x) = a x^{2} + b x + c$ if $a \neq 0$ . We call this kind of function a quadratic function.

Quadratic Function

A quadratic function, where $a$ , $b$ , and $c$ are real numbers and $a \neq 0$ , is a function of the form

$f (x) = a x^{2} + b x + c$

We graphed the quadratic function $f (x) = x^{2}$ by plotting points.

$x$	$f (x) = x^{2}$	$(x, f (x))$
$- 3$	$9$	$(- 3, 9)$
$- 2$	$4$	$(- 2, 4)$
$- 1$	$1$	$(- 1, 1)$
$0$	$0$	$(0, 0)$
$1$	$1$	$(1, 1)$
$2$	$4$	$(2, 4)$
$3$	$9$	$(3, 9)$

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 4. The y-axis of the plane runs from negative 2 to 6. The parabola has a vertex at (0, 0) and also passes through the points (-2, 4), (-1, 1), (1, 1), and (2, 4). — Figure 4.6.1

Every quadratic function has a graph that looks like this. We call this figure a parabola.

Let’s practice graphing a parabola by plotting a few points.

Example 4.6.1

Graph $f (x) = x^{2} - 1$ .

Solution

We will graph the function by plotting points.

Step 1: Choose integer values for $x$ , substitute them into the equation and simplify to find $f (x)$ .

Record the values of the ordered pairs in the chart.

$f (x) = x^{2} - 1$
$x$	$f (x)$
$0$	$- 1$
$1$	$0$
$- 1$	$0$
$2$	$3$
$- 2$	$3$

Step 2: Plot the points, and then connect them with a smooth curve. The result will be the graph of the function $f (x) = x^{2} - 1$ .

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 7 to 7. The y-axis of the plane runs from negative 4 to 10. The parabola has a vertex at (0, -1) and also passes through the points (-1, 0), (-2, 3), (1, 0), and (3, 2). — Figure 4.6.2

Try It

4) Graph $f (x) = - x^{2}$ .

Solution

This figure shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (0, 0). — Figure 4.6.3

Try It

5) Graph $f (x) = x^{2} + 1$ .

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (0, −1). — Figure 4.6.4

All graphs of quadratic functions of the form $f (x) = a x^{2} + b x + c$ are parabolas that open upward or downward. See Figure 4.6.5 below.

This image shows 2 graphs side-by-side. The graph on the left shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 2, negative 1) and passes through the points (negative 4, 3) and (0, 3). The general form for the equation of this graph is f of x equals a x squared plus b x plus c. The equation of this parabola is x squared plus 4 x plus 3. The leading coefficient, a, is greater than 0, so this parabola opens upward.The graph on the right shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (2, 7) and passes through the points (0, 3) and (4, 3). The general form for the equation of this graph is f of x equals a x squared plus b x plus c. The equation of this parabola is negative x squared plus 4 x plus 3. The leading coefficient, a, is less than 0, so this parabola opens downward. — Figure 4.6.5

Notice that the only difference in the two functions is the negative sign before the quadratic term ( $x^{2}$ ) in the equation of the graph above. When the quadratic term, is positive, the parabola opens upward, and when the quadratic term is negative, the parabola opens downward.

Parabola Orientation

For the graph of the quadratic function $f (x) = a x^{2} + b x + c$ , if:

$a > 0, the parabola opens upward$
$a < 0, the parabola opens downward$

Example 4.6.2

Determine whether each parabola opens upward or downward:

a. $f (x) = - 3 x^{2} + 2 x - 4$
b. $f (x) = 6 x^{2} + 7 x - 9$

Solution

a.

Step 1: Find the value of $a$ .

$\begin{array}{rcl} f (x) & = & a x^{2} + b x + c \\ f (x) & = & - 3 x^{2} + b x + c \\ a & = & - 3 \end{array}$

Step 2: Determine which direction the parabola will open.

Since the $a$ is negative, the parabola will open downward.

b.

Step 1: Find the value of $a$ .

$\begin{array}{rcl} f (x) & = & a x^{2} + b x + c \\ f (x) & = & 6 x^{2} + 7 x - 9 \\ a & = & 6 \end{array}$

Step 2: Determine which direction the parabola will open.

Since the $a$ is positive, the parabola will open upward.

Try It

6) Determine whether the graph of each function is a parabola that opens upward or downward:

a. $f (x) = 2 x^{2} + 5 x - 2$
b. $f (x) = - 3 x^{2} - 4 x + 7$

Solution

a. up
b. down

Try It

7) Determine whether the graph of each function is a parabola that opens upward or downward:

a. $f (x) = - 2 x^{2} - 2 x - 3$
b. $f (x) = 5 x^{2} - 2 x - 1$

Solution

a. down
b. up

Find the Axis of Symmetry and Vertex of a Parabola

Look again at Figure 4.6.5. Do you see that we could fold each parabola in half and then one side would lie on top of the other? The ‘fold line’ is a line of symmetry. We call it the axis of symmetry of the parabola.

We show the same two graphs again with the axis of symmetry. See Figure 4.6.6 below.

This image shows 2 graphs side-by-side. The graph on the left shows an upward-opening parabola and a dashed vertical line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 2, negative 1) and passes through the points (negative 4, 3) and (0, 3). The equation of this parabola is x squared plus 4 x plus 3. The vertical line passes through the point (negative 2, 0) and has the equation x equals negative 2. The graph on the right shows an downward-opening parabola and a dashed vertical line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (2, 7) and passes through the points (0, 3) and (4, 3). The equation of this parabola is negative x squared plus 4 x plus 3. The vertical line passes through the point (2, 0) and has the equation x equals 2. — Figure 4.6.6

The equation of the axis of symmetry can be derived by using the Quadratic Formula. We will omit the derivation here and proceed directly to using the result. The equation of the axis of symmetry of the graph of

$f (x) = a x^{2} + b x + c$ is $x = - \frac{b}{2 a}$

So to find the equation of symmetry of each of the parabolas we graphed above, we will substitute into the formula $x = - \frac{b}{2 a}$ .

\begin{array}{rcl} f (x) & = & a x^{2} + b x + c \\ f (x) & = & x^{2} + 4 x + 3 \\ Axis of Symmetry: \\ x & = & - \frac{b}{2 a} \\ x & = & - \frac{4}{2 \times 1} \\ x & = & - 2 \end{array}

\begin{array}{rcl} f (x) & = & a x^{2} + b x + c \\ f (x) & = & - x^{2} + 4 x + 3 \\ Axis of Symmetry: \\ x & = & - \frac{b}{2 a} \\ x & = & - \frac{4}{2 (- 1)} \\ x & = & 2 \end{array}

Notice that these are the equations of the dashed blue lines on the graphs.

The point on the parabola that is the lowest (parabola opens up), or the highest (parabola opens down), lies on the axis of symmetry. This point is called the vertex of the parabola.

We can easily find the coordinates of the vertex, because we know it is on the axis of symmetry. This means its $x$ -coordinate is $- \frac{b}{2 a}$ . To find the $y$ -coordinate of the vertex we substitute the value of the $x$ -coordinate into the quadratic function.

\begin{array}{rcl} f (x) & = & x^{2} + 4 x + 3 \\ Axis of Symmetry: x & = & - 2 \\ Vertex is (- 2,__) \\ f (x) & = & x^{2} + 4 x + 3 \\ f (x) & = & {(- 2)}^{2} + 4 (- 2) + 3 \\ f (x) & = & - 1 \\ Vertex is (- 2, - 1) \end{array}

\begin{array}{rcl} f (x) & = & - x^{2} + 4 x + 3 \\ Axis of Symmetry: x & = & 2 \\ Vertex is (2,__) \\ f (x) & = & - x^{2} + 4 x + 3 \\ f (x) & = & - {(2)}^{2} + 4 (2) + 3 \\ f (x) & = & 7 \\ Vertex is (2, 7) \end{array}

Axis of Symmetry and Vertex of a Parabola

The graph of the function $f (x) = a x^{2} + b x + c$ is a parabola where:

the axis of symmetry is the vertical line $x = - \frac{b}{2 a}$ .
the vertex is a point on the axis of symmetry, so its $x$ -coordinate is $- \frac{b}{2 a}$ .
the $y$ -coordinate of the vertex is found by substituting $x = - \frac{b}{2 a}$ into the quadratic equation.

Example 4.6.3

For the graph of $f (x) = 3 x^{2} - 6 x + 2$ find:

a. the axis of symmetry
b. the vertex

Solution

a.

$\begin{array}{rcl} f (x) & = & a x^{2} + b x + c \\ f (x) & = & 3 x^{2} - 6 x + 2 \end{array}$

Step 1: Find the axis of symmetry.

The axis of symmetry is the vertical line $x = - \frac{b}{2 a}$ .

Step 2: Substitute the values of $a, b$ into the equation.

$x = - \frac{- 6}{2 \times 3}$

Step 3: Simplify.

$x = 1$

Step 4: Write in a statement.

The axis of symmetry is the line $x = 1$ .

b.

Step 1: Find the vertex.

The vertex is a point on the line of symmetry, so its $x$ -coordinate will be $x = 1$ .

Step 2: Find $f (1)$ .

$f (1) = 3 (1)^{2} - 6 (1) + 2$

Step 3: Simplify.

$f (1) = 3 \times 1 - 6 + 2$

Step 4: The result is the $y$ -coordinate.

$f (1) = - 1$

Step 5: Write as a statement.

The vertex is $(1, - 1)$ .

Try It

8) For the graph of $f (x) = 2 x^{2} - 8 x + 1$ find:

a. the axis of symmetry
b. the vertex

Solution

a. $x = 2$
b. $(2, - 7)$

Try It

9) For the graph of $f (x) = 2 x^{2} - 4 x - 3$ find:

a. the axis of symmetry
b. the vertex

Solution

a. $x = 1$
b. $(1, - 5)$

Find the Intercepts of a Parabola

When we graphed linear equations, we often used the $x$ – and $y$ -intercepts to help us graph the lines. Finding the coordinates of the intercepts will help us to graph parabolas, too.

Remember, at the $y$ -intercept the value of $x$ is zero. So to find the $y$ -intercept, we substitute $x = 0$ into the function.

Let’s find the $y$ -intercepts of the two parabolas shown in Figure 4.6.7.

An $x$ -intercept results when the value of $f (x)$ is zero. To find an $x$ -intercept, we let $f (x) = 0$ . In other words, we will need to solve the equation $0 = a x^{2} + b x + c$ for $x$ .

$\begin{array}{rcl} (1) & f (x) & = & a x^{2} + b x + c \\ (2) & 0 & = & a x^{2} + b x + c \end{array}$

Solving quadratic equations like this is exactly what we have done earlier in this chapter!

We can now find the $x$ -intercepts of the two parabolas we looked at. First we will find the $x$ -intercepts of the parabola whose function is:

$f (x) = x^{2} + 4 x + 3$ .

Step 1: Let $f (x) = 0$ .

$0 = x^{2} + 4 x + 3$

Step 2: Factor.

$0 = (x + 1) (x + 3)$

Step 3: Solve using the Zero Product Property.

$\begin{array}{rcl} x + 1 & = & 0 \\ x & = & - 1 \\ x + 3 & = & 0 \\ x & = & - 3 \end{array}$

Step 4: Write as a statement.

The $x$ -intercepts are $(- 1, 0)$ and $(- 3, 0)$ .

Now we will find the $x$ -intercepts of the parabola whose function is:

$f (x) = - x^{2} + 4 x + 3$

Step 1: Let $f (x) = 0$ .

This quadratic does not factor, so we use the Quadratic Formula.

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Step 2: Substitute in known values.

$\begin{array}{rcl} a & = & - 1, b = 4, c = 3 \\ x & = & \frac{- 4 \pm \sqrt{4^{2} - 4 (- 1) (3)}}{2 (- 1)} \end{array}$

Step 3: Simplify.

$\begin{array}{rcl} x & = & \frac{- 4 \pm \sqrt{16 + 12}}{- 2} \\ x & = & \frac{- 4 \pm \sqrt{28}}{- 2} \\ x & = & \frac{- 4 \pm 2 \sqrt{7}}{- 2} \\ x & = & \frac{- 2 (\pm 2 \sqrt{7})}{- 2} \\ x & = & 2 \pm \sqrt{7} \end{array}$

Step 4: Write as a statement.

The $x$ -intercepts are $(2 + \sqrt{7}, 0)$ and $(2 - \sqrt{7}, 0)$ .

We will use the decimal approximations of the $x$ -intercepts, so that we can locate these points on the graph:

$\begin{array}{cccccc} (2 + \sqrt{7}, 0) \approx (4.6, 0) & (2 - \sqrt{7}, 0) \approx (- 0.6, 0) \end{array}$

Do these results agree with our graphs? See Figure 4.6.8.

This image shows 2 graphs side-by-side. The graph on the left shows the upward-opening parabola defined by the function f of x equals x squared plus 4 x plus 3 and a dashed vertical line, x equals negative 2, graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 2, negative 1). The y-intercept is (0, 3) and the x-intercepts are (negative 1, 0) and (negative 3, 0). The graph on the right shows the downward-opening parabola defined by the function f of x equals negative x squared plus 4 x plus 3 and a dashed vertical line, x equals 2, graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (2, 7). The y-intercept is (0, 3) and the x-intercepts are (2 plus square root 7, 0), approximately (4.6, 0) and (2 minus square root, 0), approximately (negative 0.6, 0). — Figure 4.6.8

Find the Intercepts of a Parabola

To find the intercepts of a parabola whose function is $f (x) = a x^{2} + b x + c$ :

$\begin{array}{cccccc} y -intercept & x -intercepts \\ Let x = 0 and solve for f (x) & Let f (x) = 0 and solve for x \end{array}$

Example 4.6.3

Find the intercepts of the parabola whose function is $f (x) = x^{2} - 2 x - 8$ .

Solution

Step 1: To find the $y$ -intercept

Let $x = 0$ and solve for $f (x)$ .

$\begin{array}{rcl} f (x) & = & x^{2} - 2 x - 8 \\ f (0) & = & 0^{2} - 2 \times 0 - 8 \\ f (0) & = & - 8 \end{array}$

When $x = 0$ , then $f (0) = - 8$ .

The $y$ -intercept is the point $(0, - 8)$ .

Step 2: To find the $x$ -intercept

Let $f (x) = 0$ and solve for $x$ .

$\begin{array}{r} f (x) = x^{2} - 2 x - 8 \\ 0 = x^{2} - 2 x - 8 \end{array}$

Step 3: Solve by factoring.

$\begin{array}{rcl} (x - 4) (x + 2) & = & 0 \\ x - 4 & = & 0 \\ x & = & 4 \\ x + 2 & = & 0 \\ x & = & - 2 \end{array}$

When $f (x) = 0$ , then $x = 4 or x = - 2$ .

The $x$ –intercepts are the points $(4, 0)$ and $(- 2, 0)$ .

Try It

10) Find the intercepts of the parabola whose function is $f (x) = x^{2} + 2 x - 8$ .

Solution

$y$ -intercept: $(0, - 8)$
$x$ -intercepts: $(- 4, 0), (2, 0)$

Try It

11) Find the intercepts of the parabola whose function is $f (x) = x^{2} - 4 x - 12$ .

Solution

$y$ -intercept: $(0, - 12)$
$x$ -intercepts: $(- 2, 0), (6, 0)$

In this chapter, we have been solving quadratic equations of the form $a x^{2} + b x + c = 0$ . We solved for $x$ and the results were the solutions to the equation.

We are now looking at quadratic functions of the form $f (x) = a x^{2} + b x + c$ . The graphs of these functions are parabolas. The $x$ -intercepts of the parabolas occur where $f (x) = 0$ .

For example:

$\begin{array}{ccccccccc} Quadratic equation & Quadratic function \\ \begin{matrix} x^{2} - 2 x - 15 & = & 0 \\ (x - 5) (x + 3) & = & 0 \\ x - 5 = 0 x + 3 & = & 0 \\ x = 5 x & = & - 3 \end{matrix} & \begin{matrix} Let f (x) = 0. \end{matrix} & \begin{array}{ccc} f (x) & = & x^{2} - 2 x - 15 \\ 0 & = & x^{2} - 2 x - 15 \\ 0 & = & (x - 5) (x + 3) \\ x - 5 & = & 0 x + 3 = 0 \\ x & = & 5 x = - 3 \end{array} \\ (5, 0) and (- 3, 0) \\ x -intercepts \end{array}$

The solutions of the quadratic function are the $x$ values of the $x$ –intercepts.

Earlier, we saw that quadratic equations have $2$ , $1$ , or $0$ solutions. The graphs below show examples of parabolas for these three cases. Since the solutions of the functions give the $x$ –intercepts of the graphs, the number of [latex]x[/latex]–intercepts is the same as the number of solutions.

Previously, we used the discriminant to determine the number of solutions of a quadratic function of the form $a x^{2} + b x + c = 0$ . Now we can use the discriminant to tell us how many $x$ -intercepts there are on the graph.

This image shows three graphs side-by-side. The graph on the left shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola lies below the x-axis and the parabola crosses the x-axis at two different points. If b squared minus 4 a c is greater than 0, then the quadratic equation a x squared plus b x plus c equals 0 has two solutions, and the graph of the parabola has 2 x-intercepts. The graph in the middle shows a downward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola lies on the x-axis, the only point of intersection between the parabola and the x-axis. If b squared minus 4 a c equals 0, then the quadratic equation a x squared plus b x plus c equals 0 has one solution, and the graph of the parabola has 1 x-intercept. The graph on the right shows an upward-opening parabola graphed on the x y-coordinate plane. The vertex of the parabola lies above the x-axis and the parabola does not cross the x-axis. If b squared minus 4 a c is less than 0, then the quadratic equation a x squared plus b x plus c equals 0 has no solutions, and the graph of the parabola has no x-intercepts. — Figure 4.6.9

Before you to find the values of the $x$ -intercepts, you may want to evaluate the discriminant so you know how many solutions to expect.

Example 4.6.4

Find the intercepts of the parabola for the function $f (x) = 5 x^{2} + x + 4$ .

Solution

Step 1: To find the $y$ -intercept

Let $x = 0$ and solve for $f (x)$ .

$\begin{array}{rcl} f (0) & = & 5 \times 0^{2} + 0 + 4 \\ f (0) & = & 4 \end{array}$

When $x = 0$ , then $f (0) = 4$ .

The $y$ -intercept is the point $(0, 4)$ .

Step 2: To find the $x$ -intercept

Let $f (x) = 0$ and solve for $x$ .

$\begin{array}{rcl} f (x) & = & 5 x^{2} + x + 4 \\ 0 & = & 5 x^{2} + x + 4 \end{array}$

Step 3: Find the value of the discriminant to predict the number of solutions which is also the number of $x$ -intercepts.

$\begin{array}{rcl} = & b^{2} - 4 a c \\ = & 1^{2} - 4 \times 5 \times 4 \\ = & 1 - 80 \\ = & - 79 \end{array}$

Since the value of the discriminant is negative, there is no real solution to the equation. There are no $x$ -intercepts.

Try It

12) Find the intercepts of the parabola whose function is $f (x) = 3 x^{2} + 4 x + 4$ .

Solution

$y$ -intercept: $(0, 4)$ no $x$ -intercept

Try It

13) Find the intercepts of the parabola whose function is $f (x) = x^{2} - 4 x - 5$ .

Solution

$y$ -intercept: $(0, - 5)$
$x$ -intercepts: $(- 1, 0), (5, 0)$

Graph Quadratic Functions Using Properties

Now we have all the pieces we need in order to graph a quadratic function. We just need to put them together. In the next example we will see how to do this.

Example 4.6.5

Graph $f (x) = x^{2} - 6 x + 8$ by using its properties.

Solution

Step 1: Determine whether the parabola opens upward or downward.

Look at $a$ in the equation.

$a = 1, = - 6, c = 8$

Since $a$ is positive, the parabola opens upward.

Step 2: Find the axis of symmetry

$f (x) = x^{2} - 6 x + 8$

The axis of symmetry is the line $x = - \frac{b}{2 a}$ .

Axis of Symmetry

$\begin{array}{rcl} x & = & - \frac{b}{2 a} \\ x & = & - \frac{(- 6)}{2 \times 1} \\ x & = & 3 \end{array}$

The axis of symmetry is the line $x = 3$ .

Step 3: Find the vertex.

The vertex is on the axis of symmetry. Substitute $x = 3$ into the function.

Vertex

$\begin{array}{rcl} f (x) & = & x^{2} - 6 x + 8 \\ f (3) & = & {(3)}^{2} - 6 (3) + 8 \\ f (3) & = & - 1 \end{array}$

The vertex is $(3, - 1)$ .

Step 4: Find the $y$ -intercept.

Find the point symmetric to the $y$ -intercept across the axis of symmetry.

We find $f (0)$ .

$y$ -intercept

$\begin{array}{rcl} f (x) & = & x^{2} - 6 x + 8 \\ f (0) & = & {(0)}^{2} - 6 (0) + 8 \\ f (0) & = & 8 \end{array}$

The $y$ -intercept is $(0, 8)$ .

We use the axis of symmetry to find a point symmetric to the $y$ -intercept. The $y$ -intercept is $3$ units left of the axis of symmetry, $x = 3$ .

A point $3$ units to the right of the axis of symmetry has $x = 6$ .

Point symmetric to $y$ -intercept: $(6, 8)$ .

Step 5: Find the $x$ -intercepts.

Find additional points if needed.

We solve $f (x) = 0$ .

We can solve this quadratic equation by factoring.

$x$ -intercepts

$\begin{array}{rcl} f (x) & = & x^{2} - 6 x + 8 \\ 0 & = & x^{2} - 6 x + 8 \\ 0 & = & (x - 2) (x - 4) \\ x & = & 2 or x = 4 \end{array}$

The $x$ -intercepts are $(2, 0)$ and $(4, 0)$ .

Step 6: Graph the parabola.

We graph the vertex, intercepts, and the point symmetric to the $y$ -intercept.

We connect these $5$ points to sketch the parabola.

An upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 2 to 10. The y-axis of the plane runs from negative 3 to 7. The parabola has a vertex at (3, negative 1). Other points plotted include the x-intercepts, (2, 0) and (4, 0), the y-intercept, (0, 8), and the point (6, 8) that is symmetric to the y-intercept across the axis of symmetry. — Figure 4.6.10

Try It

14) Graph $f (x) = x^{2} + 2 x - 8$ by using its properties.

Solution

Try It

15) Graph $f (x) = x^{2} - 8 x + 12$ by using its properties.

Solution

We list the steps to take in order to graph a quadratic function here.

HOW TO

To graph a quadratic function using properties.

Determine whether the parabola opens upward or downward.
Find the equation of the axis of symmetry.
Find the vertex.
Find the $y$ -intercept. Find the point symmetric to the $y$ -intercept across the axis of symmetry.
Find the $x$ -intercepts. Find additional points if needed.
Graph the parabola.

We were able to find the $x$ -intercepts in the last example by factoring. We find the $x$ -intercepts in the next example by factoring, too.

Example 4.6.7

Graph $f (x) = x^{2} + 6 x - 9$ by using its properties.

Solution

$\begin{array}{l} \begin{array}{c} f (x) = a x^{2} + b x + c \\ f (x) = - x^{2} + 6 x - 9 \end{array} \end{array}$

Step 1: Find the direction of the parabola.

Since a is $- 1$ , the parabola opens downward.

Step 2: Find the axis of symmetry and vertex.

To find the equation of the axis of symmetry, use $x = - \frac{b}{2 a}$ .

$\begin{array}{rcl} x & = & - \frac{b}{2 a} \\ x & = & - \frac{6}{2 (- 1)} \\ x & = & 3 \end{array}$

The axis of symmetry is $x = 3$ .

The vertex is on the line $x = 3$ .

Step 3: Graph the axis of symmetry and vertex.

A vertical line through x=3 — Figure 4.6.13

Step 4: Find $f (3)$ .

$\begin{array}{rcl} f (x) & = & - x^{2} + 6 x - 9 \\ f (3) & = & - 3^{2} + 6 \times 3 - 9 \\ f (3) & = & - 9 + 18 - 9 \\ f (3) & = & 0 \end{array}$

Step 5: Graph $f (3)$ .

The vertex is $(3, 0)$ .

A vertical line with a dot at x=3 — Figure 4.6.14

Step 6: Find $f (0)$ . The $y$ -intercept occurs when $x = 0$ .

$f (x) = - x^{2} + 6 x - 9$

Step 7: Substitute $x = 0$ .

$f (0) = - 0^{2} + 6 \times 0 - 9$

Step 8: Simplify.

$f (0) = - 9$

The $y$ -intercept is $(0, - 9)$ .

Step 9: Graph the points of symmetry.

The point $(0, - 9)$ is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is $(6, - 9)$ .

vertical line at x=3 and dots at (0,-9)(3,0)(6,-9) — Figure 4.6.15

Point symmetric to the $y$ -intercept is $(6, - 9)$

The $x$ -intercept occurs when $f (x) = 0$ .

$f (x) = - x^{2} + 6 x - 9$

Step 10: Find $f (x) = 0$ .

$0 = - x^{2} + 6 x - 9$

Step 11: Factor the GCF.

$0 = - (x^{2} - 6 x + 9)$

Step 12: Factor the trinomial.

$0 = - {(x - 3)}^{2}$

Step 13: Solve for $x$ .

$0 = 3$

Step 14: Connect the points to graph the parabola.

Parabola connecting dots (0,-9)(3,0)(6,-9) — Figure 4.6.16

Try It

16) Graph $f (x) = 3 x^{2} + 12 x - 12$ by using its properties.

Solution

This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 15 to 10. The parabola has a vertex at (2, 0). The y-intercept (0, negative 12) is plotted as well as the axis of symmetry, x equals 2. — Figure 4.6.17

Try It

17) Graph $f (x) = 4 x^{2} + 24 x + 36$ by using its properties.

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 30 to 20. The y-axis of the plane runs from negative 10 to 40. The parabola has a vertex at (negative 3, 0). The y-intercept (0, 36) is plotted as well as the axis of symmetry, x equals negative 3. — Figure 4.6.18

For the graph of $f (x) = - x^{2} + 6 x - 9$ , the vertex and the $x$ -intercept were the same point. Remember how the discriminant determines the number of solutions of a quadratic equation? The discriminant of the equation $0 = - x^{2} + 6 x - 9$ is $0$ , so there is only one solution. That means there is only one $x$ -intercept, and it is the vertex of the parabola.

How many $x$ -intercepts would you expect to see on the graph of $f (x) = x^{2} + 4 x + 5$ ?

Example 4.6.8

Graph $f (x) = x^{2} + 4 x + 5$ by using its properties.

Solution

$\begin{matrix} f (x) = a x^{2} + b x + x \\ f (x) = x^{2} + 4 x + 5 \end{matrix}$

Step 1: Find the direction of the parabola.

Since $a$ is $1$ , the parabola opens upward.

Step 2: Find the axis of symmetry.

To find the axis of symmetry, find $x = - \frac{b}{2 a}$ .

$\begin{array}{rcl} x & = & - \frac{b}{2 a} \\ x & = & - \frac{4}{(2) 1} \\ x & = & - 2 \end{array}$

Step 3: Graph the axis of symmetry.

The equation of the axis of symmetry is $x = - 2$ .

The vertex is on the line $x = - 2$ .

Step 4: Find $f (x)$ when $x = - 2$ .

$\begin{array}{rcl} f (x) & = & x^{2} + 4 x + 5 \\ f (- 2) & = & {(- 2)}^{2} + 4 (- 2) + 5 \\ f (- 2) & = & 4 - 8 + 5 \\ f (- 2) & = & 1 \end{array}$

Step 5: Graph the vertex.

The vertex is $(- 2, 1)$ .

vertical line x=-2 with dot (-2,1) — Figure 4.6.20

The $y$ -intercept occurs when $x = 0$ .

$f (x) = x^{2} + 4 x - 5$

Step 6: Find $f (0) .$

$f (0) = 5$

Step 7: Simplify.

$f (0) = 5$

The $y$ -intercept is $(0, 5)$ .

Step 8: Graph the points.

The point $(- 4, 5)$ is two units to the left of the line of symmetry.

The point two units to the right of the line of symmetry is $(0, 5)$ .

dots at (-4,5)(-2,1)(0,5) — Figure 4.6.21

Step 9: State the point that is symmetric to the $y$ -intercept.

Point symmetric to the $y$ -intercept is $(- 4, 5)$ .

The $x$ -intercept occurs when $f (x) = 0$ .

Step 10: Find $f (x) = 0$ .

$0 = x^{2} + 4 x + 5$

Step 11: Test the discriminant.

$\begin{array}{rcl} = & b^{2} - 4 a c \\ = & 4^{2} - 4 \times 1 \times 5 \\ = & 16 - 20 \\ = & - 4 \end{array}$

Since the value of the discriminant is negative, there is no real solution and so no $x$ -intercept.

Connect the points to graph the parabola. You may want to choose two more points for greater accuracy.

parabola through (-4,5)(-2,1)(0,5) — Figure 4.6.23

Try It

18) Graph $f (x) = x^{2} - 2 x + 3$ by using its properties.

Solution

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 2 to 4. The y-axis of the plane runs from negative 1 to 5. The parabola has a vertex at (1, 2). The y-intercept (0, 3) is plotted as is the line of symmetry, x equals 1. — Figure 4.6.24

Try It

19) Graph $f (x) = - 3 x^{2} - 6 x - 4$ by using its properties.

Solution

This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 2. The y-axis of the plane runs from negative 5 to 1. The parabola has a vertex at (negative 1, negative 2). The y-intercept (0, negative 4) is plotted as is the line of symmetry, x equals negative 1. — Figure 4.6.25

Finding the $y$ -intercept by finding $f (0)$ is easy, isn’t it? Sometimes we need to use the Quadratic Formula to find the $x$ -intercepts.

Example 4.6.9

Graph $f (x) = 2 x^{2} - 4 x - 3$ by using its properties.

Solution

$\begin{matrix} f (x) = a x^{2} + b x + x \\ f (x) = 2 x^{2} - 4 x - 3 \end{matrix}$

Step 1: Find the direction of the parabola.

Since $a$ is $2$ , the parabola opens upward.

Step 2: Find the axis of symmetry.

To find the equation of the axis of symmetry, use $x = - \frac{b}{2 a}$ .

$\begin{array}{rcl} x & = & - \frac{b}{2 a} \\ x & = & - \frac{- 4}{2 \times 2} \\ x & = & 1 \end{array}$

The equation of the axis of symmetry is $x = 1.$

The vertex is on the line $x = 1$ .

$f (x) = 2 x^{2} - 4 x - 3$

Step 3: Find $f (1)$ .

$\begin{array}{rcl} f (x) & = & 2 {(1)}^{2} - 4 (1) - 3 \\ f (1) & = & 2 - 4 - 3 \\ f (1) & = & - 5 \end{array}$

The vertex is $(1, - 5)$ .

The $y$ -intercept occurs when $x = 0$ .

$f (x) = 2 x^{2} - 4 x - 3$

Step 4: Find $f (0)$ .

$f (x) = 2 (0)^{2} - 4 (0) - 3$

Step 5: Simplify.

$f (0) = - 3$

The $y$ -intercept is $(0, - 3) .$

Step 6: State the point that is symmetric to the y-intercept.

The point $(0, - 3)$ is one unit to the left of the line of symmetry.

Point symmetric to the $y$ -intercept is $(2, - 3)$

The point one unit to the right of the line of symmetry is $(2, - 3)$ .

The $x$ -intercept occurs when $y = 0$ .

$f (x) = 2 x^{2} - 4 x - 3$

Step 7: Find $f (x) = 0$ .

$0 = 2 x^{2} - 4 x - 3$

Step 8: Use the Quadratic Formula.

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Step 9: Substitute in the values of $a, b,$ and $c$ .

$\begin{array}{rcl} a & = & 2, b = - 4, c = 3 \\ x & = & \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (2) (3)}}{2 (2)} \end{array}$

Step 10: Simplify.

$x = \frac{4 \pm \sqrt{16 + 24}}{4}$

Step 11: Simplify inside the radical.

$x = \frac{4 \pm \sqrt{40}}{4}$

Step 12: Simplify the radical.

$x = \frac{4 \pm 2 \sqrt{10}}{4}$

Step 13: Factor the GCF.

$x = \frac{2 (2 \pm \sqrt{10})}{4}$

Step 14: Remove common factors.

$x = \frac{2 \pm \sqrt{10}}{2}$

Step 15: Write as two equations.

$x = \frac{2 + \sqrt{10}}{2}, x = \frac{2 - \sqrt{10}}{2}$

Step 16: Approximate the values.

$x \approx 2.5, x \approx - 0.6$

The approximate values of the $x$ -intercepts are $(2.5, 0)$ and $(- 0.6, 0) .$

Step 17: Graph the parabola using the points found.

parabola through (0,-3)(1,-5)(2,-3) — Figure 4.6.26

Try It

20) Graph $f (x) = 5 x^{2} + 10 x + 3$ by using its properties.

Solution

Try It

21) Graph $f (x) = - 3 x^{2} - 6 x + 5$ by using its properties.

Solution

Solve Maximum and Minimum Applications

This figure shows 2 graphs side-by-side. The left graph shows a downward opening parabola plotted in the x y-plane. An arrow points to the vertex with the label maximum. The right graph shows an upward opening parabola plotted in the x y-plane. An arrow points to the vertex with the label minimum. — Figure 4.6.29

Knowing that the vertex of a parabola is the lowest or highest point of the parabola gives us an easy way to determine the minimum or maximum value of a quadratic function. The $y$ -coordinate of the vertex is the minimum value of a parabola that opens upward. It is the maximum value of a parabola that opens downward. See 4.6.29.

Minimum or Maximum Values of a Quadratic Function

The $y$ -coordinate of the vertex of the graph of a quadratic function is the:

- minimum value of the quadratic equation if the parabola opens upward.
- maximum value of the quadratic equation if the parabola opens downward.

Example 4.6.10

Find the minimum or maximum value of the quadratic function $f (x) = x^{2} + 2 x - 8$ .

Solution

Step 1: Find the direction of the parabola.

Since $a$ is positive, the parabola opens upward.

The quadratic equation has a minimum.

Step 2: Find the equation of the axis of symmetry.

$\begin{array}{rcl} x & = & - \frac{b}{2 a} \\ x & = & - \frac{2}{2 \times 1} \\ x & = & - 1 \end{array}$

The equation of the axis of symmetry is $x = - 1$ .

The vertex is on the line $x = - 1$ .

$f (x) = x^{2} + 2 x - 8$

Step 3: Find $f (- 1)$ .

$\begin{array}{rcl} f (- 1) & = & {(- 1)}^{2} + 2 (- 1) - 8 \\ f (- 1) & = & 1 - 2 - 8 \\ f (- 1) & = & - 9 \end{array}$

The vertex is $(- 1, - 9)$ .

Since the parabola has a minimum, the $y$ -coordinate of the vertex is the minimum $y$ -value of the quadratic equation.

The minimum value of the quadratic is $- 9$ and it occurs when $x = - 1$ .

Step 4: Show the graph to verify the result.

parabola with vertex (-1,9) — Figure 4.6.30

Try It

22) Find the maximum or minimum value of the quadratic function $f (x) = x^{2} - 8 x + 12$ .

Solution

The minimum value of the quadratic function is $- 4$ and it occurs when $x = 4$ .

Try It

23) Find the maximum or minimum value of the quadratic function $f (x) = - 4 x^{2} + 16 x - 11$ .

Solution

The maximum value of the quadratic function is $5$ and it occurs when $x = 2$ .

We have used the formula

h (t) = - 16 t^{2} + v_{0} t + h_{0}

to calculate the height in feet, $h$ , of an object shot upwards into the air with initial velocity, $v_{0}$ , after $t$ seconds .

This formula is a quadratic function, so its graph is a parabola. By solving for the coordinates of the vertex $(t, h)$ , we can find how long it will take the object to reach its maximum height. Then we can calculate the maximum height.

Example 4.6.11

The quadratic equation $h (t) = - 16 t^{2} + 176 t + 4$ models the height of a volleyball hit straight upwards with velocity $176$ feet per second from a height of 4 feet.

a. How many seconds will it take the volleyball to reach its maximum height?
b. Find the maximum height of the volleyball.

Solution

Since $a$ is negative, the parabola opens downward.

The quadratic function has a maximum.

a.

Step 1: Find the equation of the axis of symmetry.

$\begin{array}{rcl} t & = & - \frac{b}{2 a} \\ t & = & - \frac{176}{2 (- 16)} \\ t & = & 5.5 \end{array}$

Step 2: Find the axis of symmetry.

The equation of the axis of symmetry is $t = 5.5$ .

Step 3: Find the vertex.

The vertex is on the line $t = 5.5$ .

Step 4: Where is the maximum?

The maximum occurs when $t = 5.5$ seconds.

b.

Step 1: Find $h (5.5)$ .

$\begin{array}{rcl} h (t) & = & - 16 t^{2} + 176 t + 4 \\ h (t) & = & - 16 {(5.5)}^{2} + 176 (5.5) + 4 \end{array}$

Step 2: Use a calculator to simplify.

$h (t) = 488$

Step 3: State the vertex.

The vertex is $(5.5, 488)$ .

Step 4: Where is the maximum?

Since the parabola has a maximum, the $h$ -coordinate of the vertex is the maximum value of the quadratic function.

The maximum value of the quadratic is $488$ feet and it occurs when $t = 5.5$ seconds.

Step 5: State the findings.

After $5.5$ seconds, the volleyball will reach its maximum height of $488$ feet.

Try It

24) Solve, rounding answers to the nearest tenth.

The quadratic function $h (t) = - 16 t^{2} + 128 t + 32$ is used to find the height of a stone thrown upward from a height of $32$ feet at a rate of $128$ ft/sec.

How long will it take for the stone to reach its maximum height? What is the maximum height?

Solution

It will take $4$ seconds for the stone to reach its maximum height of $288$ feet.

Try It

25) A path of a toy rocket thrown upward from the ground at a rate of $208$ ft/sec is modelled by the quadratic function of $h (t) = - 16 t^{2} + 208 t$ .

When will the rocket reach its maximum height? What will be the maximum height?

Solution

It will $6.5$ seconds for the rocket to reach its maximum height of $676$ feet.

Access these online resources for additional instruction and practice with graphing quadratic functions using properties.

Key Concepts

Parabola Orientation
- For the graph of the quadratic function $f (x) = a x^{2} + b x + c$ , if
  - $a > 0$ , the parabola opens upward.
  - $a < 0$ , the parabola opens downward.
Axis of Symmetry and Vertex of a Parabola The graph of the function $f (x) = a x^{2} + b x + c$ is a parabola where:
- the axis of symmetry is the vertical line $x = - \frac{b}{2 a}$ .
- the vertex is a point on the axis of symmetry, so its $x$ -coordinate is $- \frac{b}{2 a}$ .
- the $y$ -coordinate of the vertex is found by substituting $x = - \frac{b}{2 a}$ into the quadratic equation.
Find the Intercepts of a Parabola
- To find the intercepts of a parabola whose function is $f (x) = a x^{2} + b x + c :$
  $\begin{array}{cccccc} y -intercept & x -intercepts \\ Let x = 0 and solve for f (x) & Let f (x) = 0 and solve for x \end{array}$
How to graph a quadratic function using properties.
1. Determine whether the parabola opens upward or downward.
2. Find the equation of the axis of symmetry.
3. Find the vertex.
4. Find the $y$ -intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
5. Find the $x$ -intercepts. Find additional points if needed.
6. Graph the parabola.
Minimum or Maximum Values of a Quadratic Equation
- The $y$ -coordinate of the vertex of the graph of a quadratic equation is the
- minimum value of the quadratic equation if the parabola opens upward.
- maximum value of the quadratic equation if the parabola opens downward.

Self Check

a. After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

b. After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?

Glossary

Quadratic Function: A quadratic function, where $a$ , $b$ , and $c$ are real numbers and $a \neq 0$ , is a function of the form $f (x) = a x^{2} + b x + c$ .

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Fanshawe Pre-Health Sciences Mathematics 2 Copyright © 2022 by Domenic Spilotro, MSc is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Recognize the Graph of a Quadratic Function

Quadratic Function

Parabola Orientation

Find the Axis of Symmetry and Vertex of a Parabola

Axis of Symmetry and Vertex of a Parabola

Find the Intercepts of a Parabola

Find the Intercepts of a Parabola

Graph Quadratic Functions Using Properties

HOW TO

To graph a quadratic function using properties.

Solve Maximum and Minimum Applications

Minimum or Maximum Values of a Quadratic Function

Self Check

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