30 Parabola

A quadratic function $f(x) = a x^2 + b x + c$ can be expressed in the standard form

$f(x)= a( x-h)^2 + k$

by completing the square.

The graph of $f$ is a parabola with vertex $( h, k)$ .

$\bullet$ If $a>0$ , then

$-$ the parabola opens upward and

$-$ the minimum value of $f$ is $f(h)=k$

$\bullet$ If $a<0$ , then

– the parabola opens downward and

– the maximum value of $f$ is $f(h)=k$

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Exercise 1

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The equation of the quadratic function in its standard form is :

$f(x)= -3x^2 + {30} x -{100}= -3 [x^2 - {10} x ] -{100} = -3 [(x-{5})^2 - ({5})^2 ] -{100}$

$= - 3(x-{5})^2 + 3 ({5})^2 -{100} = - 3(x-{5})^2 - ({5})^2 = - 3(x-{5})^2 - {25}$

So,

$-$ the vertex is : $( {5}, -{25})$
$-$ maximum value is $- {25}$ .

Set $y=0$ , then $- 3(x-{5})^2 - {25}= 0$ . Hence there is no $x$ -intercepts.

Set $x=0$ , then $y=-{100}$ . Hence $y=-{100}$ is the $y$ -intercept.

Exercise 2

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Set $y=0$ , then $x^2- {2} x -3= (x+1)(x- {3} )= 0$ . Hence $x=-1$ or $x= {3}$ are the $x$ -intercepts.

Set $x=0$ , then $y=-3$ . Hence $y=-3$ is the $y$ -intercept.

The equation of the quadratic function in its standard form is :

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

So,

$-$ the vertex is : $( {1}, -{4})$
$-$ the minimum value is – ${4}$ .

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Guide to Precalculus Review Copyright © 2025 by Samia CHALLAL is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

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