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# Solved Examples on the Symmetry of Functions

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(1.) Find the point that is symmetric to the point $(3, 7)$ with respect to:
(a.) $x-axis$
(b.) $y-axis$
(c.) $origin$

For each $(x, y)$ on the graph; there is also $(x, -y)$ on the same graph
Same $x$, Opposite $y$
For each $(3, 7)$; there should be $(3, -7)$
Point symmetric to $(3, 7)$ = $(3, -7)$

For each $(x, y)$ on the graph; there is also $(-x, y)$ on the same graph
Same $y$, Opposite $x$
For each $(3, 7)$; there should be $(-3, 7)$
Point symmetric to $(3, 7)$ = $(-3, 7)$

For each $(x, y)$ on the graph; there is also $(-x, -y)$ on the same graph
Opposite $x$, Opposite $y$
For each $(3, 7)$; there should be $(-3, -7)$
Point symmetric to $(3, 7)$ = $(-3, -7)$
(2.) Find the point that is symmetric to the point $(-3, 7)$ with respect to:
(a.) $x-axis$
(b.) $y-axis$
(c.) $origin$

For each $(x, y)$ on the graph; there is also $(x, -y)$ on the same graph
Same $x$, Opposite $y$
For each $(-3, 7)$; there should be $(-3, -7)$
Point symmetric to $(-3, 7)$ = $(-3, -7)$

For each $(x, y)$ on the graph; there is also $(-x, y)$ on the same graph
Same $y$, Opposite $x$
For each $(-3, 7)$; there should be $(3, 7)$
Point symmetric to $(-3, 7)$ = $(3, 7)$

For each $(x, y)$ on the graph; there is also $(-x, -y)$ on the same graph
Opposite $x$, Opposite $y$
For each $(-3, 7)$; there should be $(3, -7)$
Point symmetric to $(-3, 7)$ = $(3, -7)$
(3.) Find the point that is symmetric to the point $(3, -7)$ with respect to:
(a.) $x-axis$
(b.) $y-axis$
(c.) $origin$

For each $(x, y)$ on the graph; there is also $(x, -y)$ on the same graph
Same $x$, Opposite $y$
For each $(3, -7)$; there should be $(3, 7)$
Point symmetric to $(3, -7)$ = $(3, 7)$

For each $(x, y)$ on the graph; there is also $(-x, y)$ on the same graph
Same $y$, Opposite $x$
For each $(3, -7)$; there should be $(-3, -7)$
Point symmetric to $(3, -7)$ = $(-3, -7)$

For each $(x, y)$ on the graph; there is also $(-x, -y)$ on the same graph
Opposite $x$, Opposite $y$
For each $(3, -7)$; there should be $(-3, 7)$
Point symmetric to $(3, -7)$ = $(-3, 7)$
(4.) Find the point that is symmetric to the point $(-3, -7)$ with respect to:
(a.) $x-axis$
(b.) $y-axis$
(c.) $origin$

For each $(x, y)$ on the graph; there is also $(x, -y)$ on the same graph
Same $x$, Opposite $y$
For each $(-3, -7)$; there should be $(-3, 7)$
Point symmetric to $(-3, -7)$ = $(-3, 7)$

For each $(x, y)$ on the graph; there is also $(-x, y)$ on the same graph
Same $y$, Opposite $x$
For each $(-3, -7)$; there should be $(3, -7)$
Point symmetric to $(-3, -7)$ = $(3, -7)$

For each $(x, y)$ on the graph; there is also $(-x, -y)$ on the same graph
Opposite $x$, Opposite $y$
For each $(-3, -7)$; there should be $(3, 7)$
Point symmetric to $(-3, -7)$ = $(3, 7)$
(5.) Determine whether the function is even, odd, or neither even nor odd.
$f(x) = x\sqrt{16 - x^2}$
In terms of symmetry, what is the nature of the graph of the function?

A function is even if $f(-x) = f(x)$
A function is odd if $f(-x) = -f(x)$
If a function is not even, and is not odd; then the function is neither even nor odd

$f(x) = x\sqrt{16 - x^2} \\[3ex] f(x) = x * \sqrt{16 - x^2} \\[3ex]$ First: Use f(1) to make it easy.
$f(x) = x * \sqrt{16 - x^2} \\[3ex] f(1) = 1 * \sqrt{16 - 1^2} \\[3ex] f(1) = \sqrt{16 - 1} \\[3ex] f(1) = \sqrt{15} \\[3ex]$ Second: Test for "even"
$f(x) = x * \sqrt{16 - x^2} \\[3ex] f(-1) -1 * \sqrt{16 - (-1)^2} \\[3ex] f(-1) = -1 * \sqrt{16 - 1} \\[3ex] f(-1) = -1 * \sqrt{15} \\[3ex] f(-1) = -\sqrt{15} \\[3ex] f(-1) \ne f(1) \\[3ex] -\sqrt{15} \ne \sqrt{15} \\[3ex] Function\:\: is\:\: not\:\: even \\[3ex]$ Third: Test for "odd"
$f(x) = x * \sqrt{16 - x^2} \\[3ex] -f(1) = -1 * f(1) \\[3ex] -f(1) = -1 * \sqrt{15} \\[3ex] -f(1) = -\sqrt{15} \\[3ex] f(-1) = -f(1) \\[3ex] -\sqrt{15} = -\sqrt{15} \\[3ex] Function\:\: is\:\: odd \\[3ex]$ The graph of the function is symmetric about the $origin$
(6.) Determine whether the function is even, odd, or neither even nor odd.
$f(x) = |16x|$
In terms of symmetry, what is the nature of the graph of the function?

A function is even if $f(-x) = f(x)$
A function is odd if $f(-x) = -f(x)$
If a function is not even, and is not odd; then the function is neither even nor odd

$f(x) = |16x| \\[3ex]$ First: Use f(1) to make it easy.
$f(x) = |16x| \\[3ex] f(1) = |16 * 1| \\[3ex] f(1) = |16| \\[3ex] f(1) = 16 \\[3ex]$ Second: Test for "even"
$f(x) = |16x| \\[3ex] f(-1) = |16 * -1| \\[3ex] f(-1) = |-16| \\[3ex] f(-1) = 16 \\[3ex] f(-1) = f(1) \\[3ex] 16 = 16 \\[3ex] Function\:\: is\:\: even \\[3ex]$ The graph of the function is symmetric about the $y-axis$
(7.) Determine whether the function is even, odd, or neither even nor odd.
$f(x) = \dfrac{9x}{5x^2 - 6}$
In terms of symmetry, what is the nature of the graph of the function?

A function is even if $f(-x) = f(x)$
A function is odd if $f(-x) = -f(x)$
If a function is not even, and is not odd; then the function is neither even nor odd

$f(x) = \dfrac{9x}{5x^2 - 6} \\[3ex]$ First: Use f(1) to make it easy.
$f(x) = \dfrac{9x}{5x^2 - 6} \\[5ex] f(1) = \dfrac{9 * 1}{5(1)^2 - 6} \\[5ex] f(1) = \dfrac{9}{5(1) - 6} \\[5ex] f(1) = \dfrac{9}{5 - 6} \\[5ex] f(1) = \dfrac{9}{-1} \\[5ex] f(1) = -9 \\[3ex]$ Second: Test for "even"
$f(x) = \dfrac{9x}{5x^2 - 6} \\[5ex] f(-1) = \dfrac{9 * -1}{5(-1)^2 - 6} \\[5ex] f(1) = \dfrac{-9}{5(1) - 6} \\[5ex] f(1) = \dfrac{-9}{5 - 6} \\[5ex] f(1) = \dfrac{-9}{-1} \\[5ex] f(1) = 9 \\[3ex] f(-1) \ne f(1) \\[3ex] 9 \ne -9 \\[3ex] Function\:\: is\:\: not\:\: even \\[3ex]$ Third: Test for "odd"
$f(x) = x * \sqrt{16 - x^2} \\[3ex] -f(1) = -1 * f(1) \\[3ex] -f(1) = -1 * -9 \\[3ex] -f(1) = 9 \\[3ex] f(-1) = -f(1) \\[3ex] 9 = 9 \\[3ex] Function\:\: is\:\: odd \\[3ex]$ The graph of the function is symmetric about the $origin$
(8.) Determine whether the function is even, odd, or neither even nor odd.
$f(x) = 3x - |3x|$
In terms of symmetry, what is the nature of the graph of the function?

A function is even if $f(-x) = f(x)$
A function is odd if $f(-x) = -f(x)$
If a function is not even, and is not odd; then the function is neither even nor odd

$f(x) = 3x - |3x| \\[3ex]$ First: Use f(1) to make it easy.
$f(x) = 3x - |3x| \\[3ex] f(1) = 3(1) - |3 * 1| \\[3ex] f(1) = 3 - |3| \\[3ex] f(1) = 3 - 3 \\[3ex] f(1) = 0 \\[3ex]$ Second: Test for "even"
$f(x) = 3x - |3x| \\[3ex] f(-1) = 3(-1) - |3 * -1| \\[3ex] f(1) = -3 - |-3| \\[3ex] f(1) = -3 - 3 \\[3ex] f(1) = -6 \\[3ex] f(-1) \ne f(1) \\[3ex] -6 \ne 0 \\[3ex] Function\:\: is\:\: not\:\: even \\[3ex]$ Third: Test for "odd"
$f(x) = x * \sqrt{16 - x^2} \\[3ex] -f(1) = -1 * f(1) \\[3ex] -f(1) = -1 * 0 \\[3ex] -f(1) = 0 \\[3ex] f(-1) \ne -f(1) \\[3ex] -6 \ne 0 \\[3ex] Function\:\: is\:\: not\:\: odd \\[3ex]$ The function is neither even nor odd.
(9.) Determine whether the graph is even, odd, or neither even nor odd.
$9x^2 - 8y^2 = 4$
In terms of symmetry, what is the nature of the graph of the function?

A graph is symmetrical about the $y-axis$ if $f(-x) = f(x)$
A graph is symmetrical about the $x-axis$ if $f(x) = -f(x)$
A graph is symmetrical about the $origin$ if $f(-x) = -f(x)$

First: Let us isolate $y$
$9x^2 - 8y^2 = 4 \\[3ex] 9x^2 - 4 = 8y^2 \\[3ex] 8y^2 = 9x^2 - 4 \\[3ex] y^2 = \dfrac{9x^2 - 4}{8} \\[5ex] y = \pm \dfrac{9x^2 - 4}{8} \\[5ex] y = f(x) \\[3ex] \therefore f(x) = \pm \sqrt{\dfrac{9x^2 - 4}{8}} \\[5ex]$
This is not a function.

Teacher: Why is it not a function?
Student: It is not a function because an input value can have two output values due to the $\pm$ sign
Teacher: Correct!

Second: Use f(1) to make it easy.
$f(x) = \pm \sqrt{\dfrac{9x^2 - 4}{8}} \\[5ex] f(1) = \pm \sqrt{\dfrac{9(1)^2 - 4}{8}} \\[5ex] f(1) = \pm \sqrt{\dfrac{9(1) - 4}{8}} \\[5ex] f(1) = \pm \sqrt{\dfrac{9 - 4}{8}} \\[5ex] f(1) = \pm \sqrt{\dfrac{5}{8}} \\[5ex] f(1) = \sqrt{\dfrac{5}{8}} \:\:OR\:\: f(1) = -\sqrt{\dfrac{5}{8}} \\[5ex]$ Third: Test for "symmetry about the $y-axis$"
$f(x) = \pm \sqrt{\dfrac{9x^2 - 4}{8}} \\[5ex] f(-1) = \pm \sqrt{\dfrac{9(-1)^2 - 4}{8}} \\[5ex] f(-1) = \pm \sqrt{\dfrac{9(1) - 4}{8}} \\[5ex] f(-1) = \pm \sqrt{\dfrac{9 - 4}{8}} \\[5ex] f(-1) = \pm \sqrt{\dfrac{5}{8}} \\[5ex] f(-1) = \sqrt{\dfrac{5}{8}} \:\:OR\:\: f(1) = -\sqrt{\dfrac{5}{8}} \\[5ex] f(-1) = f(1) \\[3ex]$ The graph is symmetric about the $y-axis$

Fourth: Test for "symmetry about the $x-axis$"
$f(x) = \pm \sqrt{\dfrac{9x^2 - 4}{8}} \\[5ex] -f(1) = -1 * f(1) \\[3ex] -f(1) = -1 * \pm \sqrt{\dfrac{5}{8}} \\[5ex] -f(1) = \mp \sqrt{\dfrac{5}{8}} \\[5ex] -f(1) = -\sqrt{\dfrac{5}{8}} \:\:OR\:\: -f(1) = \sqrt{\dfrac{5}{8}} \\[5ex] f(1) = -f(1) \\[3ex]$ The graph is symmetric about the $x-axis$

Fifth: Test for "symmetry about the $origin$"
$f(x) = \pm \sqrt{\dfrac{9x^2 - 4}{8}} \\[5ex] -f(1) = -1 * f(1) \\[3ex] -f(1) = -1 * \pm \sqrt{\dfrac{5}{8}} \\[5ex] -f(1) = \mp \sqrt{\dfrac{5}{8}} \\[5ex] -f(1) = -\sqrt{\dfrac{5}{8}} \:\:OR\:\: f(1) = \sqrt{\dfrac{5}{8}} \\[5ex] f(-1) = -f(1) \\[3ex]$ The graph is symmetric about the $origin$
(10.)