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SAT II Math II : SAT Subject Test in Math II

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

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Example Questions

Example Question #1 : Graphing Quadratic Functions And Conic Sections

A baseball is thrown upward from the top of a one hundred and fifty-foot-high building. The initial speed of the ball is forty-five miles per hour. The height of the ball after seconds is modeled by the function

$h (t) = -16t^{2}+ 45t +150$

How high does the ball get (nearest foot)?

Possible Answers:

$182 \textrm{ ft}$

$205 \textrm{ ft}$

$195 \textrm{ ft}$

$173 \textrm{ ft}$

$166 \textrm{ ft}$

Correct answer:

$182 \textrm{ ft}$

Explanation:

A quadratic function such as has a parabola as its graph. The high point of the parabola - the vertex - is what we are looking for.

The vertex of a function

$h (t) = at^{2}+bt +c$

has as coordinates

$\left ( - \frac{b}{2a} , h \left ( - \frac{b}{2a} \right ) \right )$ .

The second coordinate is the height and we are looking for this quantity. Since $h (t) = -16t^{2}+ 45t +150$ , setting a = -16, b= 45 :

$- \frac{b}{2a} = - \frac{45}{2(-16)} = \frac{45}{32} = 1.40625$ seconds for the ball to reach the peak.

The height of the ball at this point is h ( 1.40625 ) , which can be evaluated by substitution:

$h (t) = -16t^{2}+ 45t +150$

$h (1.40625 ) = -16 (1.40625 )^{2}+ 45 (1.40625 ) +150$

$\approx -16 (1.9775 ) + 45 (1.40625 ) +150$

$\approx -31.6406 + 63.2813 + 150$

$\approx 181.6407$

Round this to 182 feet.

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Example Question #71 : Functions And Graphs

Give the -intercept(s) of the parabola of the equation

$f(x) = 3x^{2}-9x -54$

Possible Answers:

(-54,0)

(-3,0) and (6,0)

$\left ( -\frac{3}{2}, 0 \right )$ and $\left ( \frac{3}{2}, 0 \right )$

The parabola has no -intercept.

(-4,0) and (9,0)

Correct answer:

(-3,0) and (6,0)

Explanation:

Set f(x)= 0 and solve for :

$f(x) = 3x^{2}-9x -54$

$3x^{2}-9x -54 = 0$

The terms have a GCF of 2, so

$3\left ( x^{2}- 3x - 18 \right )= 0$

The trinomial in parentheses can be FOILed out by noting that -6 +3 = -3 and $-6 \times 3= -18$ :

3(x-6)(x+3)=0

And you can divide both sides by 3 to get rid of the coefficient:

(x-6)(x+3)=0

Set each of the linear binomials to 0 and solve for :

$x - 6 = 0 \Rightarrow x = 6$

$x +3= 0 \Rightarrow x = -3$

The parabola has as its two intercepts the points (-3,0) and (6,0) .

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Example Question #81 : Functions And Graphs

Give the amplitude of the graph of the function

$f (x) = 2 \pi \sin 2 \pi x$

Possible Answers:

$\pi$

$4 \pi$

$2 \pi$

Correct answer:

$2 \pi$

Explanation:

The amplitude of the graph of a sine function $f(x) = A \sin Bx$ is . Here, $A= 2 \pi$ , so this is the amplitude.

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Example Question #2 : Graphing Functions

Which of these functions has a graph with amplitude 4?

Possible Answers:

$f(x) = 9\cos 4x$

$f(x) = 4 \cos \frac{\pi x}{5}$

$f(x) = \frac{1}{7}\cos \frac{\pi x }{4}$

$f(x) = \frac{1}{4}\cos \frac{\pi x}{3}$

$f(x) = \frac{2}{3}\cos \frac{ x }{4}$

Correct answer:

$f(x) = 4 \cos \frac{\pi x}{5}$

Explanation:

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

$f(x)= A \cos Nx$ .

The graph of a cosine function in this form has amplitude . Therefore, for this function to have amplitude 4, A = 4 . Of the five choices, only

$f(x) = 4 \cos \frac{\pi x}{5}$

matches this description.

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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 #51 : Trigonometric Functions And Graphs

Which of the given functions has the greatest amplitude?

Possible Answers:

$\sin(4x)$

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

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

$2\sin(x)$

$\sin(x+3)$

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

None

Two

Three

Four

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 #2 : Graphing Piecewise And Recusive 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:

Three

Four

None

Two

One

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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All SAT II Math II Resources

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SAT II Math II : SAT Subject Test in Math II

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

All SAT II Math II Resources

Example Questions

Example Question #1 : Graphing Quadratic Functions And Conic Sections

Example Question #71 : Functions And Graphs

Example Question #81 : Functions And Graphs

Example Question #2 : Graphing Functions

Example Question #2 : Graphing Trigonometric Functions

Example Question #4 : Graphing Trigonometric Functions

Example Question #51 : Trigonometric Functions And Graphs

Example Question #1 : Graphing Piecewise And Recusive Functions

Example Question #2 : Graphing Piecewise And Recusive Functions

Example Question #3 : Graphing Piecewise And Recusive Functions

All SAT II Math II Resources

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