A graph is a subset of the plane

The graph of an equation in $$x$$ and $$y$$ is the set of all points with coordinates $$(x,y)$$ satisfying the equation. In the case of a function $$y = f ( x )$$, the graph of $$f$$ is the point set $$\{ (x,y)\, | \,y = f ( x ) \}$$.

Plotting Points

The most fundamental way to sketch the graph of a function is to plot points.

1. Select an interval on the $$x$$-axis encompassing all or part of the domain of the function.
2. Select a finite set of $$x$$ values in the domain of the function lying in that selected interval.
3. For each of the selected values of $$x$$, compute the corresponding value of $$y$$ using the formula for $$y = f ( x )$$.
4. Plot this finite set of points $$(x,y)$$ in the plane and sketch the graph by connecting the points, being careful not to connect two adjacent points unless all the input values lying between the two points are in the domain of the function.

Shortcuts

There are certain shortcuts which may make the task of sketching graphs easier by reducing the number of points which must be plotted. These shortcuts are

1. symmetry
2. shifting, i.e. translation
3. stretching or shrinking
4. reflecting

These shortcuts apply not only to the graphs of functions (vertical line test) but to all graphs having an equation in $$x$$ and $$y$$.

Symmetry with respect to the $$y$$-axis

A graph is symmetric with respect to the $$y$$-axis if the point $$( - x, y )$$ is on the graph whenever the point $$( x, y )$$ is on the graph. This means that if one replaces every occurrence of $$x$$ in the equation with $$( - x )$$ [ be sure to put parentheses around the $$- x$$ ] and then simplifies the equation, the resulting equation will be equivalent to the original equation.

Exercise 1.7.1

Let $$f(x)=4-x^2$$ on the interval $$[ - 3, 3 ]$$. Replace the $$x$$ in the equation with $$( - x )$$ and simplify. Did you get the same equation? Sketch the graph and notice that for every point $$( x, y )$$ on the graph, the mirror image of that point in the $$y$$-axis, namely the point $$( - x, y )$$, is also on the graph.

See Solution

Even functions

A function which is symmetric with respect to the $$y$$-axis is called an even function and has the special property that $$f ( - x ) = f ( x )$$ for every $$x$$ in the domain of the function.

Exercise 1.7.2

Show that the function $$f(x)=x^4-5x^2+3$$ is an even function.

See Solution

Symmetry with respect to the $$x$$-axis

A graph is symmetric with respect to the $$x$$-axis if the point $$( x, - y )$$ is on the graph whenever the point $$( x, y )$$ is on the graph.

This means that if one replaces every occurrence of $$y$$ in the equation with $$- y )$$ [ be sure to put parentheses around the $$- y$$ ] and then simplifies the equation, the resulting equation will be equivalent to the original equation.

Note that if $$y$$ is not 0 then $$( x, y )$$ and $$( x, - y )$$ are two points which lie on the same vertical line, so the graph would not be the graph of a function in that case. [ Why not? ]

Exercise 1.7.3

Let $$x=y^2-1$$ for $$y$$ in the interval $$[ - 2, 2 ]$$. Replace the $$y$$ in the equation with $$( - y )$$ and simplify. Did you get the same equation? Sketch the graph and notice that for every point $$( x, y )$$ on the graph, the mirror image of that point in the $$x$$-axis, namely the point $$( x, - y )$$, is also on the graph.

See Solution

Symmetry with respect to the origin

A graph is symmetric with respect to the origin if the point $$( - x, - y )$$ is on the graph whenever the point $$( x, y )$$ is on the graph.

This means that if one replaces every occurrence of $$x$$ in the equation with $$( - x )$$ and every occurrence of $$y$$ in the equation with $$( - y )$$ [ be sure to put parentheses around the $$- x$$ and $$- y$$ ] and then simplifies the equation, the resulting equation will be equivalent to the original equation.

Notice that the point $$( - x, - y )$$ is the same distance from the origin as the point $$( x, y )$$ , except it is on the opposite side of the origin and that the line connecting the points $$( x, y )$$ and $$( - x, - y )$$ always passes through the origin $$(0,0)$$.

Exercise 1.7.4

Let $$y=x^3$$ for x in the interval $$[ - 2, 2 ]$$. Replace the $$x$$ in the equation with $$( - x )$$ and replace the $$y$$ in the equation with $$( - y )$$ and simplify. Did you get the same equation? Sketch the graph and notice that for every point $$( x, y )$$ on the graph, the point $$( - x, - y )$$, is also on the graph.

See Solution

Odd Functions

A function which is symmetric with respect to the origin is called an odd function and has the special property that $$f ( - x ) = - f ( x )$$ for every $$x$$ in the domain of the function.

Exercise 1.7.5

Show that the function $$f(x)=x^3-2x$$ is an odd function.

See Solution

By noticing the symmetry of a graph, one need plot only half as many points, obtaining the other points of the graph by symmetry.

Horizontal translation

In mathematics, to translate a geometrical figure means to change it's position without distorting or rotating it.

The horizontal translation principle states that replacing each occurrence of $$x$$ in an $$x y$$ equation with $$( x - h )$$ [ be sure to put parentheses around the $$x - h$$ ] produces an equation whose graph lies $$h$$ units horizontally from the graph of the original equation. The $$h$$ can be any real number, positive or negative.

Exercise 1.7.6

Sketch the graph of $$y=x^2$$, $$y=(x-1)^2$$ and $$y=(x+1)^2$$. Notice the application of the horizontal translation principle.

See Solution

Horizontal translation and function notation

Stated in terms of function notation, the horizontal translation principle states that the graph of $$y = f ( x - h )$$ is shifted $$h$$ units horizontally from the graph of $$y = f ( x )$$.

Thus the graph of $$y = f ( x - 2 )$$ will lie two units to the right of the graph of $$y = f ( x )$$ and the graph of $$y = f ( x + 3 ) = f ( x - ( - 3 ) )$$ will lie three units to the left of y = f ( x ). That is, minus $$3$$ units means three units to the left.

Exercise 1.7.7

Sketch the graph of $$f(x)=\sqrt{x}$$. Then sketch the graphs of $$g(x)=f(x-2)=\sqrt{x-2}$$ and $$p(x)=f(x+1)=\sqrt{x+1}$$.

See Solution

Vertical Translation

The vertical translation principle states that replacing each occurrence of $$y$$ in an $$x y$$ equation with $$( y - k )$$ [ be sure to put parentheses around the $$y - k$$ ] produces an equation whose graph lies $$k$$ units vertically from the graph of the original equation. The $$k$$ can be any real number, positive or negative.

Exercise 1.7.8

Sketch the graph of $$y=x^2$$, $$y-1=x^2$$ and $$y+1=x^2$$. Notice the application of the vertical translation principle.

See Solution

Vertical translation and function notation

Stated in terms of function notation, the vertical translation principle states that the graph of $$y - k = f ( x )$$ [ alternately, $$y = f ( x ) + k$$ ] is shifted $$k$$ units vertically from the graph of $$y = f ( x )$$.

Thus the graph of $$y - 2 = f ( x )$$ will lie two units above the graph of $$y = f ( x )$$ and the graph of $$y + 3= f ( x )$$ will lie three units below $$y = f ( x )$$.

Exercise 1.7.9

Sketch the graph of $$f(x)=\sqrt{x}$$.Then sketch the graphs of $$g(x)=\sqrt{x}+2$$ and $$p(x)=\sqrt{x}-1$$.

See Solution

Combining horizontal and vertical translations

Both the horizontal and vertical translation principles may be applied to the same problem.

Given an equation in $$x$$ and $$y$$ if each occurrence of $$x$$ is replaced by $$(x-h)$$ and each occurrence of $$y$$ is replaced by $$(y-k)$$ then the resulting equation will have a graph equivalent to shifting the graph of the original equation $$h$$ units horizontally and $$k$$ units vertically. Stated in function notation, if the original function is $$y=g(x)$$ then the equation of the translated function will be $$y-k=g(x-h)$$. Solving for $$y$$ gives $$y=g(x-h)+k$$ as the equation for the new function. If we call the new function $$f$$ then we have $$f(x)=g(x-h)+k$$. So the graph of $$f$$ is the graph of $$g$$ shifted $$h$$ units horizontally and $$k$$ units vertically.

Exercise 1.7.10

Sketch the graph of $$f(x)=\sqrt{x-2}+1$$ using both the horizontal and vertical translation principles.

See Solution

Horizontal Stretching and shrinking

The horizontal stretching/shrinking principle states that replacing each occurrence of $$x$$ in an $$x y$$ equation with $$\left(\dfrac{x}{c}\right)$$ stretches the graph horizontally by a factor of $$c$$ when $$c > 1$$ and shrinks the graph horizontally by a factor $$c$$ when $$0 < c < 1$$.

Exercise 1.7.11

Sketch the graph of $$y=\left(\dfrac{x}{2}\right)^2$$ together with the graph of $$y=x^2$$. Notice that the first graph is twice as wide as the second.

See Solution

Exercise 1.7.12

Sketch the graph of $$y=(2x)^2=\left(\dfrac{x}{1/2}\right)^2$$ together with the graph of $$y=x^2$$. Notice that the first graph is only one-half as wide as the second.

See Solution

Vertical stretching and shrinking

The vertical stretching/shrinking principle states that replacing each occurrence of $$y$$ in an $$x y$$ equation with $$\left(\dfrac{y}{c}\right)$$ stretches the graph vertically by a factor of $$c$$ when $$c > 1$$ and shrinks the graph vertically by a factor $$c$$ when $$0 < c < 1$$.

Exercise 1.7.13

Sketch the graph of $$\dfrac{y}{2}=x^2$$ [ after first re-writing as $$y=2x^2$$ ] together with the graph of $$y=x^2$$. Notice that the first graph is twice as tall as the second.

See Solution

Exercise 1.7.14

Sketch the graph of $$f(x)=4x^2$$ together with the graph of $$g(x)=x^2$$. Compare this problem to exercise 1.7.12.

Notice that this particular problem can be worked either by vertical stretching or by horizontal shrinking!

See Solution

Exercise 1.7.15

Sketch the graph of $$y=\dfrac{x^2}{2}$$ [ that is, $$2y=x^2$$ or $$\dfrac{y}{1/2}=x^2$$ ], together with the graph of $$y=x^2$$. Notice that the first graph is only one-half as tall as the second.

See Solution

Exercise 1.7.16

Sketch the graph of $$f(x)=\left(\dfrac{x}{\sqrt{2}}\right)^2$$ together with the graph of $$g(x)=x^2$$. Hint: Compare to exercise 1.7.15. Notice that the graph of $$f(x)=\left(\dfrac{x}{\sqrt{2}}\right)^2$$ is wider than the graph of $$g(x)=x^2$$.

See Solution

Vertical reflection

The vertical reflection principle states that replacing each occurrence of $$x$$ in an $$x y$$ equation with $$( - x )$$ [ be sure to put parentheses around the $$- x$$ ] reflects the graph of the original equation about the $$\mathbf{y}$$-axis.

Exercise 1.7.17

Sketch the graphs of $$y=\sqrt{x}$$ and $$y=\sqrt{-x}$$.

See Solution

Horizontal reflection

The horizontal reflection principle states that replacing each occurrence of $$y$$ in an $$x y$$ equation with $$( - y )$$ [ be sure to put parentheses around the $$- y$$ ] reflects the graph of the original equation about the $$\mathbf{x}$$-axis.

Exercise 1.7.18

Sketch the graphs of $$y=\sqrt{x}$$ and $$-y=\sqrt{x}$$ [ which is equivalent to $$y=-\sqrt{x}$$ ]. Note how this example differs from Exercise 1.7.17 above.

See Solution

Exercise 1.7.19

Sketch the graph of $$f(x)=-\sqrt{1-x}$$ using the principles of reflection and translation.

Hint: First, graph $$-y=\sqrt{-x}$$ and re-write the equation for $$f(x)$$ in the equivalent form $$-y=\sqrt{-(x-1)}$$

See Solution