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SAT II Math I : Functions and Graphs

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

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Example Questions

Example Question #41 : Solving Functions

Solve:

$\frac{3}{x+1}+\frac{4}{x^2-1}=0$

Possible Answers:

$x=\frac{5}{3}$

x=0

$x=-\frac{5}{3}$

x=3

$x=-\frac{1}{3}$

Correct answer:

$x=-\frac{1}{3}$

Explanation:

To solve this equation, we must find the least common denominator to add the fractions.

Keep in mind that the denominator of the second term is the difference of squares, which can be rewritten as

(x^2-1)=(x+1)(x-1)

This is the least common denominator.

Now, we multiply both sides of the equationby the LCD on top and bottom (this is essentially 1):

$\frac{(x+1)(x-1)}{(x+1)(x-1)}\cdot (\frac{3}{x+1}+\frac{4}{x^2-1})=0$

After canceling terms, we get

$\frac{3(x-1)+4}{x^2-1}=0$

Now, we solve for x:

3(x-1)+4=0

$x=-\frac{1}{3}$

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Example Question #1 : Adding And Subtracting Fractions

Simplify $\frac{2}{3}+\frac{1}{4}+\frac{5}{6}$

Possible Answers:

$\frac{8}{13}$

$1\frac{3}{4}$

$1\frac{1}{3}$

$1\frac{1}{6}$

$\frac{11}{12}$

Correct answer:

$1\frac{3}{4}$

Explanation:

Find the least common denominator (LCD) and convert each fraction to the LCD and then add. Simplify as necessary.

$\frac{2}{3}+\frac{1}{4}+\frac{5}{6}=\frac{8}{12}+\frac{3}{12}+\frac{10}{12}=\frac{21}{12}$

The result is an improper fraction because the numerator is larger than the denominator and can be simplified and converted to a mix numeral.

$\frac{21}{12}=\frac{7}{4}=1\frac{3}{4}$

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Example Question #33 : Operations With Fractions

Find the simplified result:

$\frac{5}{6} + \frac{7}{8}$

Possible Answers:

$\frac{40}{48}$

$\frac{82}{48}$

$\frac{42}{48}$

$\frac{12}{14}$

$\frac{41}{24}$

Correct answer:

$\frac{41}{24}$

Explanation:

Start by making both fractions into the same denominator. One option is $6 \times 8 = 48$

Then adjust the numerators by multiplying each fraction's numerator by the other fraction's denominator:

$\frac{5}{6}\cdot \frac{8}{8}+ \frac{7}{8}\cdot \frac{6}{6}$

$=\frac{40}{48}+\frac{42}{48}$

Then add the adjusted numerators:

$=\frac{82}{48}$

Then we simplify by dividing both numerator and denominator by 2:

$=\frac{2 \cdot 41}{2 \cdot 24}=\frac{41}{24}$

which gives us the final result.

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Example Question #8 : Solving Rational And Fractional Functions

Simplify:

$\frac{25st^2}{6s^3t^5}\div\frac{30s^{-9}}{45s^2t^7}$

Possible Answers:

$\frac{25}{4s^9t^4}$

$\frac{4}{25s^2t}$

$\frac{25s^9t^4}{4}$

$\frac{4s^9t^4}{25}$

Correct answer:

$\frac{25s^9t^4}{4}$

Explanation:

To divid fractions, you need to multiply the first fraction by the reciprocal of the second.

$(\frac{25st^2}{6s^3t^5})\div(\frac{30s^{-9}}{45s^2t^7})=(\frac{25st^2}{6s^3t^5})\times(\frac{45s^2t^7}{30s^{-9}})$

Now, multiply the numerators and denominators together, then simplify.

$(\frac{25st^2}{6s^3t^5})\times(\frac{45s^2t^7}{30s^{-9}})=\frac{1125s^3t^9}{180s^{-6}t^5}=\frac{25s^9t^4}{4}$

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Example Question #2 : Multiplying And Dividing Fractions

Simplify:

$\frac{12m^2n}{16m^3n^4}\div\frac{8m^{-4}n}{18m^4n^2}$

Possible Answers:

$\frac{27m^7}{16n^2}$

$\frac{16n^2}{27m^7}$

27m^7n^2

$\frac{27n^2}{16m^7}$

Correct answer:

$\frac{27m^7}{16n^2}$

Explanation:

To divide fractions, you need to multiply the first fraction by the reciprocal of the second.

$(\frac{12m^2n}{16m^3n^4})\div(\frac{8m^{-4}n}{18m^4n^2})=(\frac{12m^2n}{16m^3n^4})\times(\frac{18m^4n^2}{8m^{-4}n})$

Now, multiply the numerators and denominators, then simplify.

$(\frac{12m^2n}{16m^3n^4})\times(\frac{18m^4n^2}{8m^{-4}n})=\frac{216m^6n^3}{128m^{-1}n^5}=\frac{27m^7}{16n^2}$

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Example Question #91 : Functions And Graphs

Solve: $\frac{1}{\sqrt{3x+1}} = \frac{1}{3}$

Possible Answers:

$\frac{8}{5}$

$\frac{1}{4}$

$\frac{1}{2}$

$\frac{8}{3}$

Correct answer:

$\frac{8}{3}$

Explanation:

Cross multiply to eliminate the fractions.

$(1)\sqrt{3x+1} = (1)(3)$

$\sqrt{3x+1} =3$

Square both sides.

$(\sqrt{3x+1})^2 =3^2$

3x+1 =9

Subtract 1 on both sides.

3x+1 -1=9-1

3x=8

Divide by 3 on both sides.

$\frac{3x}{3}=\frac{8}{3}$

The answer is: $\frac{8}{3}$

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Example Question #92 : Functions And Graphs

A baseball is thrown straight up with an initial speed of 50 feet per second by a man standing on the roof of a 120-foot high building. The height of the baseball in feet, as a function of time in seconds , is modeled by the function

$h(t) = -16t^{2}+50t + 120$

To the nearest tenth of a second, how long does it take for the baseball to hit the ground?

Possible Answers:

$1.6\textrm{ sec}$

$7.5 \textrm{ sec}$

$6.4\textrm{ sec}$

$3.1\textrm{ sec}$

$4.7\textrm{ sec}$

Correct answer:

$4.7\textrm{ sec}$

Explanation:

When the baseball hits the ground, the height is 0, so we set $h\left ( t\right ) = 0$ . and solve for .

$-16t^{2}+50t + 120 = 0$

This can be done using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Set a = -16, b = 50, c = 120 :

$t = \frac{-50 \pm \sqrt{50^{2}-4(-16)120}}{2(-16)}$

$t = \frac{-50 \pm \sqrt{2,500- (- 7,680)}}{-32}$

$t = \frac{-50 \pm \sqrt{2,500+7,680}}{-32}$

$t = \frac{-50 \pm \sqrt{10,180}}{-32}$

$t \approx \frac{-50 \pm 100.9}{-32}$

One possible solution:

$t \approx \frac{-50 +100.9}{-32} \approx -1.6$

We throw this out, since time must be positive.

The other:

$t \approx \frac{-50 -100.9}{-32} \approx 4.7$

This solution, we keep. The baseball hits the ground in about 4.7 seconds.

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Example Question #93 : Functions And Graphs

Define $f (x) = x^{2} - 5$ and g (x) = 2x + 7 .

Find $(g \circ f) (x)$

Possible Answers:

$(g \circ f) (x) = 2 x^{2}-3$

$(g \circ f) (x) =4 x ^{2} + 28 x + 54$

$(g \circ f) (x) = 2x^{3} +7x^{2} - 10x - 35$

$(g \circ f) (x) = 2 x^{2}-17$

$(g \circ f) (x) =4 x ^{2} + 28 x +44$

Correct answer:

$(g \circ f) (x) = 2 x^{2}-3$

Explanation:

By definition, $(g \circ f) (x) = g (f(x))$ , so

$(g \circ f) (x) = g ( x^{2} - 5 )$

$= 2 (x^{2}-5 ) + 7$

$= 2 x^{2}-10 + 7$

$= 2 x^{2}-3$

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Example Question #1 : Finding Roots

$y=x^{2}-2x-3$

Factor the above function to find the roots of the quadratic equation.

Possible Answers:

x=-3,1

x=-3,-2

x=-1,3

x=2,3

Correct answer:

x=-1,3

Explanation:

Factoring a quadratic equation means doing FOIL backwards. Recall that when you use FOIL, you start with two binomials and end with a trinomial:

$(x+a)(x+b)=x^{2}+(a+b)x+ab$

Now, we're trying to go the other direction -- starting with a trinomial, and going back to two factors.

Here, -3 is equal to , and -2 is equal to a+b . We can use this information to find out what and are, separately. In other words, we have to find two factors of -3 that add up to -2.

Factors of -3:

3*-1 (sum = 2)
-3*1 (sum = -2)

Thus our factored equation should look like this:

y=(x-3)(x+1)

The roots of the quadratic equation are the values of x for which y is 0.

0=(x-3)(x+1)

We know that anything times zero is zero. So the entire expression equals zero when at least one of the factors equals zero.

$x-3=0\: \: \textup{and} \: \: x+1=0$

$x=3 \: \: \textup{and}\: \: -1$

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Example Question #1 : Finding Roots

Find the roots of the function:
$\small f(x)=x^2+4x-12$

Possible Answers:

$\small \small x=2\ or\ -6$

$\small \small x=-2\ or\ 6$

$\small x=-3\ or\ 4$

$\small x=-4\ or\ 3$

Correct answer:

$\small \small x=2\ or\ -6$

Explanation:

Factor:

0=x^2+4x-12

$\small \small 0=(x-2)(x+6)$

Double check by factoring:

$\small F: (x)(x)$

$\small \small O:(x)(6)=6x$

$\small \small I: (-2)(x)=-2x$

$\small \small L: (-2)(6)=-12$

Add together: $\small x^2+6x-2x-12=x^2+4x-12$

Therefore:

$\small x-2=0\ or x+6=0$

$\small x=2\or-6$

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SAT II Math I : Functions and Graphs

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

All SAT II Math I Resources

Example Questions

Example Question #41 : Solving Functions

Example Question #1 : Adding And Subtracting Fractions

Example Question #33 : Operations With Fractions

Example Question #8 : Solving Rational And Fractional Functions

Example Question #2 : Multiplying And Dividing Fractions

Example Question #91 : Functions And Graphs

Example Question #92 : Functions And Graphs

Example Question #93 : Functions And Graphs

Example Question #1 : Finding Roots

Example Question #1 : Finding Roots

All SAT II Math I Resources

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