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SAT II Math II : Graphing Functions

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Example Questions

Example Question #1 : Graphing Functions

Circle

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?

Possible Answers:

Insufficient information is given to determine the equation of the line.

$y = - \frac{4}{3} x+ 16$

$y = \frac{4}{3} x$

$y = \frac{3}{4} x+ \frac{7}{2}$

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

Correct answer:

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

Explanation:

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 (0,0),(6,8) , 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 (6,8) , 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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Example Question #1 : Pre Calculus

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

(x-2)^2+y^2=36

Possible Answers:

$(-2,0),\ r=6$

$(-2,0),\ r=36$

$(2,0),\ r=36$

$(2,0),\ r=6$

Correct answer:

$(2,0),\ r=6$

Explanation:

A circle is defined by an equation in the format (x-h)^2+(y-k)^2=r^2 .

The center is indicated by the point (h,k) and the radius .

In the equation (x-2)^2+(y)^2=36=6^2 , the center is (2,0) and the radius is .

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Example Question #1 : Graphing Quadratic Functions And Conic Sections

Give the axis of symmetry of the parabola of the equation

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

Possible Answers:

x = - 16

$x= \frac{1}{6}$

$x= -\frac{1}{6}$

$x = -\frac{3}{2}$

$x = \frac{3}{2}$

Correct answer:

$x = -\frac{3}{2}$

Explanation:

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 a = 3, b = 9 :

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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Example Question #2 : Graphing Quadratic Functions And Conic Sections

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

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

Possible Answers:

129

Correct answer:

Explanation:

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

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

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

Substitute a= 4, b = -16 :

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

The -coordinate is therefore f(2) :

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

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

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

f(2) = 16 - 32 + 1

f(2) = -15

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Example Question #3 : Graphing Quadratic Functions And Conic Sections

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?

Possible Answers:

$140\textrm{ ft}$

$132\textrm{ ft}$

$156 \textrm{ ft}$

$120\textrm{ ft}$

$172\textrm{ ft}$

Correct answer:

$156 \textrm{ ft}$

Explanation:

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 a = -16, b = 60 :

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

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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Example Question #2 : Graphing Functions

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

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

Possible Answers:

$\frac{37}{2}$

$\frac{1}{2}$

$\frac{3}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

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

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

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

Set a = 2, b = 6 :

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

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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Example Question #291 : Sat Subject Test In Math Ii

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:

$173 \textrm{ ft}$

$205 \textrm{ ft}$

$166 \textrm{ ft}$

$195 \textrm{ ft}$

$182 \textrm{ ft}$

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 #4 : Graphing Quadratic Functions And Conic Sections

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

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

Possible Answers:

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

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

The parabola has no -intercept.

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

(-54,0)

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 #1 : Graphing Trigonometric Functions

Give the amplitude of the graph of the function

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

Possible Answers:

$4 \pi$

$\pi$

$2 \pi$

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 Trigonometric Functions

Which of these functions has a graph with amplitude 4?

Possible Answers:

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

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

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

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

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

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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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 #1 : Graphing Functions

Example Question #1 : Pre Calculus

Example Question #1 : Graphing Quadratic Functions And Conic Sections

Example Question #2 : Graphing Quadratic Functions And Conic Sections

Example Question #3 : Graphing Quadratic Functions And Conic Sections

Example Question #2 : Graphing Functions

Example Question #291 : Sat Subject Test In Math Ii

Example Question #4 : Graphing Quadratic Functions And Conic Sections

Example Question #1 : Graphing Trigonometric Functions

Example Question #2 : Graphing Trigonometric Functions

All SAT II Math II Resources

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