Functions and Graphs - SAT Subject Test in Math II

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure. The circle has its center at the origin; the line is tangent to the circle at the point indicated. What is the equation of the line in slope-intercept form?

Answer

A line tangent to a circle at a given point is perpendicular to the radius from the center to that point. That radius, which has endpoints , has slope

$m = \frac{8-0}{6-0} = \frac{8}{6} = \frac{4}{3}$ .

The line, being perpendicular to this radius, will have slope equal to the opposite of the reciprocal of that of the radius. This slope will be $-\frac{3}{4}$ . Since it includes point , we can use the point-slope form of the line to find its equation:

$y - y_{1} = m(x-x_{1})$

$y - 8= - \frac{3}{}4\left ( x-6 \right )$

$y - 8= - \frac{3}{4} x+ \frac{9}{2}$

$y - 8+ 8 = - \frac{3}{4} x+ \frac{9}{2} + 8$

$y = - \frac{3}{4} x+ \frac{25}{2}$

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Question

Give the period of the graph of the equation

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

Answer

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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Question

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)

III) $x= \pi$

Answer

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 :

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

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

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

Since the values coincide, is continuous at .

$1-\tan \frac{\pi }{2}$ 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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Question

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?

Answer

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 $(-\infty, -5]$

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

or

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

on the interval

However, this value is not in the interval , so this is not an -intercept.

on the interval

However, this value is not in the interval , so this is not an -intercept.

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

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

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

The graph of has no -intercepts.

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Question

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?

Answer

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.

on the interval $(-\infty, -5]$

However, this value is not in the interval , so this is not an -intercept.

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

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

or

is on the interval , so is an -intercept.

$g(x)= x^{2}-x- 6$ on the interval

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

is on the interval , so is an -intercept.

on the interval $(5, \infty)$

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

The graph has two -intercepts, and .

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Question

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)

II)

III)

Answer

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

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

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

evaluated for :

Since the values coincide, the graph of is continuous at .

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

evaluated for :

evaluated for :

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

evaluated for :

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

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

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

II and III only is the correct response.

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Question

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.

Answer

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

on the interval ,

evaluate:

The -intercept is .

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Question

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.

Answer

To find the -intercept, evaluate 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 .

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Question

What is the center and radius of the circle indicated by the equation?

Answer

A circle is defined by an equation in the format .

The center is indicated by the point and the radius .

In the equation , the center is and the radius is .

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Question

Give the -coordinate of the vertex of the parabola of the function

$f(x) = 2x^{2} + 6x + 5$ .

Answer

The -coordinate of the vertex of a parabola of the form

$f(x) = ax^{2}+ bx + c$

is

$x= -\frac{b}{2a}$ .

Set :

$x= -\frac{6}{2\cdot 2} = -\frac{6}{4} =- \frac{3}{2}$

The -coordinate is therefore $f\left (- \frac{3}{2} \right )$ :

$f(x) = 2x^{2} + 6x + 5$

$f \left ( -\frac{3}{2} \right )= 2\left ( -\frac{3}{2} \right )^{2}+ 6\left ( -\frac{3}{2} \right ) + 5$

$= 2\left ( \frac{9}{4} \right ) + 6\left (- \frac{3}{2} \right ) + 5$

$= \frac{9}{2} - 9 + 5 = \frac{1}{2}$ , which is the correct choice.

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Question

Give the axis of symmetry of the parabola of the equation

$f(x) = 3x^{2} + 9x - 16$

Answer

The line of symmetry of the parabola of the equation

$f(x) = ax^{2}+ bx + c$

is the vertical line

$x = -\frac{b}{2a}$

Substitute :

The line of symmetry is

$x = -\frac{9}{2 \cdot 3} = - \frac{9}{6} = -\frac{3}{2}$

That is, the line of the equation $x =- \frac{3}{2}$ .

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Question

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

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

Answer

Set 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 and $-6 \times 3= -18$ :

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

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

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

or

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

The parabola has as its two intercepts the points and .

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Question

Give the -coordinate of the vertex of the parabola of the function

$f(x) = 4x^{2}- 16x + 1$

Answer

The -coordinate of the vertex of a parabola of the form

$f(x) = ax^{2}+ bx + c$

is

$x= -\frac{b}{2a}$ .

Substitute :

$x= -\frac{-16}{2\cdot 4} = -\frac{-16}{8} = 2$

The -coordinate is therefore :

$f(x) = 4x^{2}- 16x + 1$

$f(2) = 4 \cdot 2 ^{2}- 16 \cdot 2 + 1$

$f(2) = 4 \cdot 4- 16 \cdot 2 + 1$

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Question

A baseball is thrown straight up with an initial speed of 60 miles per hour by a man standing on the roof of a 100-foot high building. The height of the baseball in feet is modeled by the function

$h(t) = -16t^{2}+60t + 100$

To the nearest foot, how high is the baseball when it reaches the highest point of its path?

Answer

We are seeking the value of when the graph of $h\left ( t\right )$ - a parabola - reaches its vertex.

To find this value, we first find the value of . For a parabola of the equation

$f\left ( t\right ) = at^{2}+ bt + c$ ,

the value of the vertex is

$t = -\frac{b}{2a}$ .

Substitute :

$t = -\frac{60}{2 (-16)} = 1.875$

The height of the baseball after 1.875 seconds will be

$h(1.875) = -16 \cdot 1.875^{2}+60 \cdot 1.875 + 100$

$h(1.875) = -56.25 +112.5 + 100 \approx 156$ feet.

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Question

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)?

Answer

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 :

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

The height of the ball at this point is , 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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Question

Which of the given functions has the greatest amplitude?

Answer

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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Question

Which of these functions has a graph with amplitude 4?

Answer

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, . Of the five choices, only

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

matches this description.

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Question

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

Answer

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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Question

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

Answer

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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Question

Give the amplitude of the graph of the function

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

Answer

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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