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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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All SAT II Math I Resources

6 Diagnostic Tests 113 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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Example Question #2 : Graphing Other Functions

$f(x)=5x^{5}-3x^{2}+7$

Where does f(x) cross the axis?

Possible Answers:

-3

-7

Correct answer:

Explanation:

f(x) crosses the axis when equals 0. So, substitute in 0 for :

$f(x)=5x^{5}-3x^{2}+7$

$f(0)=5(0)^{5}-3(0)^{2}+7$

f(0)=7

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Example Question #3 : Graphing Other Functions

Screen_shot_2014-12-24_at_2.27.32_pm

Which of the following is an equation for the above parabola?

Possible Answers:

$y=\frac{1}{2}(x+3)(x-2)$

$y=\frac{1}{2}(x-3)(x+2)$

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

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

Correct answer:

$y=\frac{1}{2}(x-3)(x+2)$

Explanation:

The zeros of the parabola are at and , so when placed into the formula

$y=C(x-z_{1})(x-z_{2})$ ,

each of their signs is reversed to end up with the correct sign in the answer. The coefficient can be found by plugging in any easily-identifiable, non-zero point to the above formula. For example, we can plug in (4,3) which gives

3=C(4-3)(4+2)

6C=3

$C=\frac{1}{2}$

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Example Question #1 : Graphing Polynomial Functions

Which equation best represents the following graph?

Graph6

Possible Answers:

None of these

y(x)=-x-1

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

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

$y(x)=e^{2x-1}+\frac{1}{9}$

Correct answer:

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

Explanation:

We have the following answer choices.

$y(x)=e^{2x-1}+\frac{1}{9}$

The first equation is a cubic function, which produces a function similar to the graph. The second equation is quadratic and thus, a parabola. The graph does not look like a prabola, so the 2nd equation will be incorrect. The third equation describes a line, but the graph is not linear; the third equation is incorrect. The fourth equation is incorrect because it is an exponential, and the graph is not an exponential. So that leaves the first equation as the best possible choice.

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Example Question #1 : Graphing Polynomial Functions

Which of the graphs best represents the following function?

$f(x)=\frac{9}{10}x^2-7x+2$

Possible Answers:

Graph_cube_

Graph_line_

None of these

Graph_parabola_

Graph_exponential_

Correct answer:

Graph_parabola_

Explanation:

$f(x)=\frac{9}{10}x^2-7x+2$

The highest exponent of the variable term is two ( x^2 ). This tells that this function is quadratic, meaning that it is a parabola.

The graph below will be the answer, as it shows a parabolic curve.

Graph_parabola_

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Example Question #4 : Graphing Other Functions

Which of the following is a graph for the following equation:

x^8+x^7-x^4+10

Possible Answers:

Cannot be determined

Incorrect 3

Correct answer

Incorrect 2

Incorrect 1

Correct answer:

Correct answer

Explanation:

The way to figure out this problem is by understanding behavior of polynomials.

The sign that occurs before the x^8 is positive and therefore it is understood that the function will open upwards. the "8" on the function is an even number which means that the function is going to be u-shaped. The only answer choice that fits both these criteria is:

Correct answer

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Example Question #1 : Graphing Other Functions

Define a function $f(x) = x^{5}+ 4x^{2}$ .

f(c) = 7 for exactly one real value of on the interval (1,0. 1.5) .

Which of the following statements is correct about ?

Possible Answers:

$c \in (1.0, 1.1 )$

$c \in ( 1.3, 1.5)$

$c \in ( 1.3, 1.4)$

$c \in (1.1, 1.2)$

$c \in ( 1.2, 1.3)$

Correct answer:

$c \in (1.1, 1.2)$

Explanation:

Define $g(x)= f(x) - 7 = x^{5}+ 4x^{2} -7$ . Then, if g(c)= f(c) - 7 = 0 , it follows that f(c) = 7 .

By the Intermediate Value Theorem (IVT), if g(x) is a continuous function, and g(a) and g(b) are of unlike sign, then g(c) = 0 for some $c \in (a, b)$ . As a polynomial, g(x) is a continuous function, so the IVT applies here.

Evaluate g(x) for each of the following values: $\left \{ 1.0, 1.1, 1.2, 1.3, 1.4, 1.5\right \}$

$g(1.0) = 1.0^{5}+ 4 \cdot 1.0^{2} -7 = -2$

$g(1.1) = 1.1^{5}+ 4 \cdot 1.1^{2} -7 \approx -0.55$

$g(1.2) = 1.2^{5}+ 4 \cdot 1.2^{2} -7 \approx 1.24$

$g(1.3) = 1.3^{5}+ 4 \cdot 1.3^{2} -7 \approx 3.47$

$g(1.4) = 1.4^{5}+ 4 \cdot 1.4^{2} -7 \approx 6.22$

$g(1.5) = 1.5^{5}+ 4 \cdot 1.5^{2} -7 \approx 9.59$

Only in the case of (1.1, 1.2) does it hold that g (x) assumes a different sign at both endpoints - g(1.1) < 0 < g(1.2) . By the IVT, g(c) = 0 , and f(c) = 7 , for some $c \in (1.1, 1.2)$ .

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Example Question #1 : Solving Functions From Word Problems

At Joe's pizzeria a pizza costs $5 with the first topping, and then an additional 75 cents for each additional topping.

If represents the number of toppings on a pizza, what function represents the cost of a pizza with at least one topping?

Possible Answers:

f(x) = 5x + 0.75

f(x) = 5 + 75x

f(x) = 0.75 + 5(x+1)

f(x) = 5 + 75(x-1)

f(x) = 0.75(x - 1)+5

Correct answer:

f(x) = 0.75(x - 1)+5

Explanation:

Notice that the question describes a linear equation because there is a constant rate of change (the cost per topping). This means we can use slope intercept form to describe the scenario.

Recall that slope intercept form is

The value of is the -value when x = 0 . In this case x=0 means there are zero additional toppings and the question tells us that this pizza costs $5 so:

5=m(0) +b

5=b

is the slope, or rate of change as increases. In this case each additional topping costs cents. Notice that the units have shifted from dollars to cents so we must write this as 0.75

m=0.75

Notice that the first topping is included in the $5 cost. The cent charge only applies to additional toppings. So one less than the number of toppings is equivalent to:

x-1

Putting all these steps together we get:

0.75 (x-1) +5

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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 #2 : Graphing Other Functions

Example Question #3 : Graphing Other Functions

Example Question #1 : Graphing Polynomial Functions

Example Question #1 : Graphing Polynomial Functions

Example Question #4 : Graphing Other Functions

Example Question #1 : Graphing Other Functions

Example Question #1 : Solving Functions From Word Problems

All SAT II Math I Resources

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