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SAT II Math II : Graphing Functions

Study concepts, example questions & explanations for SAT II Math II

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Example Questions

Example Question #2 : Graphing Trigonometric Functions

Which of these functions has a graph with amplitude $\frac{4}{7}$ ?

Possible Answers:

$f(x)=\sin \frac{4x}{7}$

$f(x)=\sin \frac{4 \pi x}{7}$

$f(x)=\sin \frac{7x}{4}$

$f(x)= \frac{4}{7} \sin x$

$f(x)= \frac{2}{7} \sin x$

Correct answer:

$f(x)= \frac{4}{7} \sin x$

Explanation:

The functions in each of the choices take the form of a sine function

$f(x)= A \sin Bx$ .

The graph of a sine function in this form has amplitude . Therefore, for this function to have amplitude 4, $A = \frac{4}{7}$ . Of the five choices, only

$f(x)= \frac{4}{7} \sin x$

matches this description.

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Example Question #4 : Graphing Trigonometric Functions

Which of the following sine functions has a graph with period of 7?

Possible Answers:

$f(x)= \frac{11}{5} \sin \frac{7 x}{2}$

$f(x)= \frac{11}{5} \sin \frac{2 \pi x}{7}$

$f(x)= \frac{11}{5} \sin \frac{7 \pi x}{2}$

$f(x)= \frac{11}{5} \sin \frac{ \pi x}{7}$

$f(x)= \frac{11}{5} \sin \frac{2 x}{7}$

Correct answer:

$f(x)= \frac{11}{5} \sin \frac{2 \pi x}{7}$

Explanation:

The period of the graph of a sine function $f(x)= A \sin B x$ , is $\frac{2 \pi}{B}$ , or $2 \pi \div B$ .

Therefore, we solve for :

$\frac{2 \pi}{B} = 7$

$\frac{B}{2 \pi} = \frac{1}{7}$

$B= \frac{1}{7} \cdot 2 \pi = \frac{2 \pi}{7}$

The correct choice is therefore $f(x)= \frac{11}{5} \sin \frac{2 \pi x}{7}$ .

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Example Question #2 : Period And Amplitude

Which of the given functions has the greatest amplitude?

Possible Answers:

$\sin(4x)$

$\sin(x+3)$

$2\sin(x)$

$\sin\frac{2\pi }{3}(x+2)$

$\frac{1}{2}\sin(4x)$

Correct answer:

$2\sin(x)$

Explanation:

The amplitude of a function is the amount by which the graph of the function travels above and below its midline. When graphing a sine function, the value of the amplitude is equivalent to the value of the coefficient of the sine. Similarly, the coefficient associated with the x-value is related to the function's period. The largest coefficient associated with the sine in the provided functions is 2; therefore the correct answer is $2\sin(x)$ .

The amplitude is dictated by the coefficient of the trigonometric function. In this case, all of the other functions have a coefficient of one or one-half.

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Example Question #1 : Graphing Piecewise And Recusive Functions

Define a function as follows:

$g(x) = \left\{\begin{matrix} x^{2}-6x-7 &x\le -5 \\ 6-x & -5 < x \le 0 \\ x+6 &0< x \le 5\\ x^{2}-x- 6& x> 5 \end{matrix}\right.$

How many -intercept(s) does the graph of have?

Possible Answers:

One

Two

Four

None

Three

Correct answer:

None

Explanation:

To find the -coordinates of possible -intercepts, set each of the expressions in the definition equal to 0, making sure that the solution is on the interval on which is so defined.

$g(x)= x^{2}-6x-7$ on the interval $(-\infty, -5]$

g(x) = 0

$x^{2}-6x-7 = 0$

(x+1)(x-7)= 0

x= -1 or x= 7

However, neither value is in the interval $(-\infty, -5]$ , so neither is an -intercept.

g(x)= 6-x on the interval ( -5, 0]

g(x) = 0

6-x = 0

$x = 6 \;$

However, this value is not in the interval ( -5, 0] , so this is not an -intercept.

g(x) =0 on the interval (0, 5]

x+ 6 = 0

x = -6

However, this value is not in the interval (0, 5] , so this is not an -intercept.

$g(x)= x^{2}-x- 6$ on the interval $(5, \infty)$

g(x)= 0

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

(x+2) (x- 3)= 0

x= -2,x= 3

However, neither value is in the interval $(5, \infty)$ , so neither is an -intercept.

The graph of has no -intercepts.

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Example Question #11 : Graphing Functions

Define a function as follows:

$g(x) = \left\{\begin{matrix} 6-x &x\le -5 \\ x^{2}-6x-7 & -5 < x \le 0 \\x^{2}-x- 6 &0< x \le 5\\x+6 & x> 5 \end{matrix}\right.$

How many -intercept(s) does the graph of have?

Possible Answers:

Four

One

None

Two

Three

Correct answer:

Two

Explanation:

To find the -coordinates of possible -intercepts, set each of the expressions in the definition equal to 0, making sure that the solution is on the interval on which is so defined.

g(x)= 6-x on the interval $(-\infty, -5]$

g(x) = 0

6-x = 0

$x = 6 \;$

However, this value is not in the interval ( -5, 0] , so this is not an -intercept.

$g(x)= x^{2}-6x-7$ on the interval ( -5, 0]

g(x) = 0

$x^{2}-6x-7 = 0$

(x+1)(x-7)= 0

x= -1 or x= 7

x= -1 is on the interval ( -5, 0] , so (-1,0) is an -intercept.

$g(x)= x^{2}-x- 6$ on the interval (0, 5]

g(x)= 0

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

(x+2) (x- 3)= 0

x= -2,x= 3

x=3 is on the interval (0, 5] , so (3,0) is an -intercept.

g(x) =0 on the interval $(5, \infty)$

x+ 6 = 0

x = -6

However, this value is not in the interval $(5, \infty)$ , so this is not an -intercept.

The graph has two -intercepts, (-1,0) and (3,0) .

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Example Question #3 : Graphing Piecewise And Recusive Functions

Define function as follows:

$g(x) = \left\{\begin{matrix} x^{2}-2x+7 &x\le -1 \\ 9-x & -1 < x \le 0 \\ x+8 &0< x \le 1 \\ x^{2}+3x+ 6& x> 1 \end{matrix}\right.$

Give the -intercept of the graph of the function.

Possible Answers:

(0, 9)

(0, 7)

(0, 6)

(0, 8)

The graph does not have a -intercept.

Correct answer:

(0, 9)

Explanation:

To find the -intercept, evaluate g(0) using the definition of on the interval that includes the value 0. Since

g(x)= 9-x

on the interval (-1, 0] ,

evaluate:

g(0)= 9-0 = 9

The -intercept is (0, 9) .

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Example Question #3 : Graphing Piecewise And Recusive Functions

Define a function as follows:

$g(x) = \left\{\begin{matrix} \sin \frac{x}{2} -1 &x\le - \pi \\ \cos 2x & -\pi < x \le 0 \\ 1-\tan \frac{x }{2}&0< x < \pi \\ 0 & x\ge \pi \end{matrix}\right.$

At which of the following values of is discontinuous?

I) $x = -\pi$

II) x= 0

III) $x= \pi$

Possible Answers:

I and III only

All of I, II, and III

II and III only

I and II only

None of I, II, and III

Correct answer:

I and III only

Explanation:

To determine whether is continuous at $x = -\pi$ , we examine the definitions of on both sides of $-\pi$ , and evaluate both for $x = -\pi$ :

$\sin \frac{x}{2} -1$ evaluated for $x = -\pi$ :

$\sin \frac{-\pi}{2} -1 = \sin \left (- \frac{ \pi}{2} \right )-1= -1-1= -2$

$\cos 2x$ evaluated for $x = -\pi$ :

$\cos 2(-\pi) = \cos (-2\pi)= 1$

Since the values do not coincide, is discontinuous at $x = -\pi$ .

We do the same thing with the other two boundary values 0 and $\pi$ .

$\cos 2x$ evaluated for x =0 :

$\cos 2(0) = \cos 0= 1$

$1-\tan \frac{x }{2}$ evaluated for x =0 :

$1-\tan \frac{0 }{2} = 1 - \tan 0= 1-0 = 1$

Since the values coincide, is continuous at x =0 .

$1-\tan \frac{\pi }{2}$ turns out to be undefined for $x = \pi$ , (since $\tan \frac{\pi }{2}$ is undefined), so is discontinuous at $x = \pi$ .

The correct response is I and III only.

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Example Question #5 : Graphing Piecewise And Recusive Functions

Define a function as follows:

$g(x) = \left\{\begin{matrix} x^{2}-2x+7 &x\le -1 \\ 9-x & -1 < x \le 0 \\ x+8 &0< x \le 1 \\ x^{2}+3x+ 6& x> 1 \end{matrix}\right.$

At which of the following values of is the graph of discontinuous?

I) x = -1

II) x= 0

III) x= 1

Possible Answers:

I and III only

I and II only

None of I, II, and III

All of I, II, and III

II and III only

Correct answer:

II and III only

Explanation:

To determine whether is continuous at x = -1 , we examine the definitions of on both sides of , and evaluate both for x = -1 :

$x^{2}-2x+7$ evaluated for x = -1 :

$(-1)^{2}-2(-1)+7= 1+2+7 = 10$

9-x evaluated for x = -1 :

9-(-1)= 10

Since the values coincide, the graph of is continuous at x = -1 .

We do the same thing with the other two boundary values 0 and 1:

9-x evaluated for x = 0 :

9-0= 9

x+8 evaluated for x = 0 :

0+8 = 8

Since the values do not coincide, the graph of is discontinuous at x = 0 .

x+8 evaluated for x = 1 :

1+8 = 9

$x^{2}+3x+ 6$ evaluate for x = 1 :

$1^{2}+3\cdot 1+ 6 = 1+ 3 + 6 = 10$

Since the values do not coincide, the graph of is discontinuous at x = 1 .

II and III only is the correct response.

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Example Question #1 : Graphing Piecewise And Recusive Functions

Define a function as follows:

$g(x) = \left\{\begin{matrix} \sin \frac{x}{2} -1 &x\le - \pi \\ \cos 2x & -\pi < x \le 0 \\ 1-\tan \frac{x }{2}&0< x < \pi \\ 0 & x\ge \pi \end{matrix}\right.$

Give the -intercept of the graph of the function.

Possible Answers:

The graph does not have a -intercept.

(0,0)

(0, -2)

(0, -1)

(0, 1)

Correct answer:

(0, 1)

Explanation:

To find the -intercept, evaluate g(0) using the definition of on the interval that includes the value 0. Since

$g(x)= \cos 2x$

on the interval $(-\pi, 0]$ ,

evaluate:

$g(x)= \cos 2 \cdot 0 = \cos 0 = 1$

The -intercept is (0, 1) .

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Example Question #1 : Graphing Parametric Functions

Give the period of the graph of the equation

$f(x) = \frac{4}{7} \cos \left ( \frac{3\pi x}{7} \right )$

Possible Answers:

$\frac{7}{3}$

$\frac{8}{7}$

$\frac{2}{7}$

$\frac{14}{3}$

$\frac{4}{7}$

Correct answer:

$\frac{14}{3}$

Explanation:

The period of the graph of a cosine function $f(x)= A \cos Nx$ is $\frac{2 \pi}{N}$ , or $2 \pi \div N$

Since $N = \frac{3\pi}{7}$ , the period is

$2 \pi \div \frac{3\pi}{7} = \frac{2 \pi }{1}\times \frac{7} {3\pi}= \frac{14}{3}$

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SAT II Math II : Graphing Functions

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #2 : Graphing Trigonometric Functions

Example Question #4 : Graphing Trigonometric Functions

Example Question #2 : Period And Amplitude

Example Question #1 : Graphing Piecewise And Recusive Functions

Example Question #11 : Graphing Functions

Example Question #3 : Graphing Piecewise And Recusive Functions

Example Question #3 : Graphing Piecewise And Recusive Functions

Example Question #5 : Graphing Piecewise And Recusive Functions

Example Question #1 : Graphing Piecewise And Recusive Functions

Example Question #1 : Graphing Parametric Functions

All SAT II Math II Resources

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