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29 Graphs

Plane curve:   is the graph  of an equation in two variables; that is the set of all points (x,y) that satisfy the equation.


In particular, the graph of a function f is  the set     

                                   G_f=\{ (x,y):  \,\, y= f(x) \,\, \hbox{ and  } \,\,x\in D_f\}

                x-Intercepts = x-coordinates of a point where the graph crosses the x-axis.

                y-Intercepts = y-coordinates of a point where the graph crosses the y-axis.

 

Symmetry :    a graph G  is


- symmetric with respect to the y-axis if  :  \qquad \,\, (a,b) \in G  \quad \Longleftrightarrow\quad (-a, b)\in G


- symmetric with respect to the x-axis if  :  \qquad \,\, (a,b) \in G  \quad \Longleftrightarrow\quad (a, -b)\in G


- symmetric with respect to  the origin if  :  \qquad  \,\, (a,b) \in G  \quad \Longleftrightarrow\quad (-a, -b)\in G


- symmetric with respect to the line y= x  if  :  \qquad \,\, (a,b) \in G  \quad \Longleftrightarrow\quad (b, a)\in G

 

The vertical line test:  A plane curve is the graph of a function if and only if no vertical line intersects the curve more than once.

 

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Exercise 1

 

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Figure A is the graph of a function because the vertical line rule is satisfied.
Figure B is not the graph of a function because the vertical line rule is not satisfied.
Figure C is the graph of a function because the vertical line rule is satisfied.
Figure D cannot be the graph of a function because the vertical line x = 1 intersects the curve into two different points   (1, y_1)  and  (1, y_2).
Figure E is the graph of a function because the vertical line rule is satisfied.
Figure F is not the graph of a function because the vertical line rule is not satisfied.

 

Exercise 2

 

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Set x=0, then y^2={25}. Hence y=\pm {5} are the y-intercepts.

Set y=0 , then \displaystyle{ x^2=\frac{{25}}{{25}}=1 }. Hence \displaystyle{ x=\pm 1 are the x-intercepts.

 

Exercise 3

 

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Set y=0, then {3} | x|={12}. Hence \displaystyle{ x=\pm \frac{{12}}{{3}} = \pm 4 } are the x-intercepts.

Set x=0 , then y=-{12}. Hence y=-{12} is the y-intercept.

 

Exercise 4

 

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* Substitute -y for y : {a} x^2 + (- y)^2 = {a} x^2 + y^2 = {b}.

Since the equation is unchanged, the graph has x-axis symmetry.

* Substitute -x for x : {a} (-x)^2 + y^2= {a} x^2 + y^2 = {b}.
Since the equation is unchanged, the graph has y-axis symmetry.

* Substitute -x for x and -y for y : {a} (-x)^2 +(- y)^2 = {a} x^2 + y^2 = {b}.

Since the equation is unchanged, the graph has origin symmetry.

 

Exercise 5

 

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* Substitute -y for y : -y = {5} x - {6}.

Since the equation is changed, the graph does not have x-axis symmetry.

* Substitute -x for x : y={5} (-x) - {6} = - {5} x - {6}.
Since the equation is changed, the graph does not have y-axis symmetry.

* Substitute -x for x and -y for y : -y = - {5} x - {6} \Leftrightarrow y = {5} x + {6}.

Since the equation is changed, the graph does not have origin symmetry.

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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.