Properties of Functions and Graphs - SAT Subject Test in Math II

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Question

What is the vertex of ? Is it a max or min?

Answer

The polynomial is in standard form of a parabola.

To determine the vertex, first write the formula.

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

Substitute the coefficients.

$x=-\frac{-5}{2(-3)} = -\frac{5}{6}$

Since the is negative is negative, the parabola opens down, and we will have a maximum.

The answer is: $-\frac{5}{6}, \textup{Max}$

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Question

Given the parabola equation , what is the max or minimum, and where?

Answer

The parabola is in the form:

The vertex formula will determine the x-value of the max or min. Since the value of is negative, the parabola will open downward, and there will be a maximum.

Write the vertex formula and substitute the correct coefficients.

$x=-\frac{b}{2a} = -\frac{4}{2(-3)} = \frac{2}{3}$

Substitute this value back in the parabolic equation to determine the y-value.

$y=-3( \frac{2}{3})^2+4( \frac{2}{3}) = \frac{4}{3}$

The answer is: $\textup{Max}, (\frac{2}{3}, \frac{4}{3})$

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Question

Define the functions and on the set of real numbers as follows:

$f (x) = \sqrt{x- 16}$

$g(x) = \sqrt{x- 4}$

Give the natural domain of the composite function $f \circ g$ .

Answer

The natural domain of the composite function $f \circ g$ is defined to be the intersection of the two sets.

One set is the natural domain of . Since is defined to be the square root of an expression, the radicand must be nonnegative. Therefore,

$x - 4 \geq 0$

$x \geq 4$

This set is $[4, \infty)$ .

The other set is the set of numbers that the function pairs with a number within the domain of . Since the radicand of the square root in must be nonnegative,

$x - 16 \geq 0$

$x \geq 16$

For to fall within this set:

$g(x) \geq 16$

$\sqrt{x- 4} \geq 16$

$x - 4 \geq 256$

$x \geq 260$

This set is $[260, \infty)$ .

$[4, \infty) \cap [260, \infty) = [260, \infty)$ , which is the natural domain.

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Question

Define $f(x) = \sqrt[3]{100 - x^{2}}$

Give the domain of .

Answer

Every real number has one real cube root, so there are no restrictions on the radicand of a cube root expression. The domain is the set of all real numbers.

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Question

Define $f (x) = \sqrt{16-x^{2}}$ .

Give the range of .

Answer

The radicand within a square root symbol must be nonnegative, so

$16 - x^{2} \geq 0$

$16 \geq x^{2}$

$x^{2} \leq 16$

This happens if and only if $-4 \leq x \leq 4$ , so the domain of is .

$f (x) = \sqrt{16-x^{2}}$ assumes its greatest value when $16-x^{2}$ , which is the point on where $x^{2}$ is least - this is at .

$f (0) = \sqrt{16-0^{2}} = \sqrt{16} = 4$

Similarly, $f (x) = \sqrt{16-x^{2}}$ assumes its least value when $16-x^{2}$ , which is the point on where $x^{2}$ is greatest - this is at $x = \pm 4$ .

$f (4) = \sqrt{16-4^{2}} = \sqrt{0} = 0$

$f (-4) = \sqrt{16- \left (-4 \right )^{2}} = \sqrt{0} = 0$

Therefore, the range of is .

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Question

Define $f(x)= \frac{x- 4}{x-7}$ .

Give the domain of .

Answer

In a rational function, the domain excludes exactly the value(s) of the variable which make the denominator equal to 0. Set the denominator to find these values:

The domain is the set of all real numbers except 7 - that is, $(-\infty, 7) \cup (7, \infty)$ .

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Question

Define $f(x)= \frac{x- 4}{x-7}$ .

Give the range of .

Answer

The function can be rewritten as follows:

$f(x)= \frac{x- 4}{x-7}$

$f(x)= \frac{x- 7 + 3}{x-7}$

$f(x)= \frac{x- 7 }{x-7}+ \frac{ 3}{x-7}$

$f(x)=1+ \frac{ 3}{x-7}$

The expression $\frac{ 3}{x-7}$ can assume any value except for 0, so the expression $f(x)=1+ \frac{ 3}{x-7}$ can assume any value except for 1. The range is therefore the set of all real numbers except for 1, or

$(-\infty, 1) \cup (1, \infty)$ .

This choice is not among the responses.

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Question

Define $f(x) = 2- 4\cos x$

Give the range of .

Answer

$-1 \leq \cos x \leq 1$ for any real value of .

Therefore,

$-4(-1) \cdot \geq -4 \cdot \cos x \geq -4 \cdot 1$

$4 \geq -4 \cos x \geq -4$

$4 + 2 \geq -4 \cos x + 2 \geq -4 + 2$

$6 \geq 2-4 \cos x \geq -2$

$-2 \leq f(x) \leq 6$

The range is .

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Question

Define $f(x) = 2- \cos 4x$

Give the range of .

Answer

$-1 \leq \cos \theta \leq 1$ for any real value of $\theta$ , so

$-1 \leq \cos 4x \leq 1$

$-1 \cdot \left (-1 \right ) \geq -1 \cdot \cos 4x \geq -1 \cdot 1$

$1 \geq - \cos 4x \geq -1$

$1+ 2 \geq - \cos 4x + 2 \geq -1 + 2$

$3 \geq 2 - \cos 4x \geq 1$

$1 \leq f(x) \leq 3$ ,

making the range $\left [ 1,3\right ]$ .

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Question

Define $f(x) = \frac{5}{\sin 4x}$

Give the range of .

Answer

$f(x) = \frac{5}{\sin 4x}$ can be rewritten as $f (x) = 5 \csc 4x$ .

For all real values of $\theta$ ,

$\csc \theta\leq -1$ or $\csc \theta\geq 1$ .

Therefore,

$\csc 4x \leq -1$ or $\csc 4x\geq 1$ and

$5 \csc 4x \leq -5$ or $5 \csc 4x\geq 5$ .

The range of is $(- \infty, -5] \cup [5, \infty)$ .

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Question

What is the domain of the function

Answer

The domain of a function is all the x-values that in that function. The function is a upward facing parabola with a vertex as (0,3). The parabola keeps getting wider and is not bounded by any x-values so it will continue forever. Parenthesis are used because infinity is not a definable number and so it can not be included.

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Question

What is the domain of the function? $y=\sqrt{-10x+3}$

Answer

Notice this function resembles the parent function $y=\sqrt {x}$ . The value of must be zero or greater.

Set up an inequality to determine the domain of .

$-10x+3\geq 0$

Subtract three from both sides.

$-10x+3-3\geq 0-3$

$-10x\geq -3$

Divide by negative ten on both sides. The sign will switch.

$\frac{-10x}{-10}\geq \frac{-3}{-10}$

The domain is: $x\leq \frac{3}{10}$

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Question

What is the range of the function $f(x)=\frac{2}{x}-3$ ?

Answer

Start by considering the term $\frac{2}{x}$ . $\frac{2}{x}$ will hold for all values of , except when . Thus, $\frac{2}{x}-3$ must be defined by all values except since the equation is just shifted down by .

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Question

What is the range of the equation $y=\frac{1}{5}$ ?

Answer

The equation given represents a horizontal line. This means that every y-value on the domain is equal to $\frac{1}{5}$ .

The answer is: $[\frac{1}{5}]$

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Question

What is the slopeof the line between the points (-1,0) and (3,5)?

Answer

For this problem we will need to use the slope equation:

$m=\frac{y_2-y_1}{x_2-x_1}$

In our case and

Therefore, our slope equation would read:

$m=\frac{5-0}{3--1}=\frac{5}{4}$

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Question

What is the slope of the function

Answer

To find the slope of this function we first need to get it into slope-intercept form

where

To do this we need to divide the function by 3:

From here we can see our m, which is our slope equals 2

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Question

What is the slope for the line having the following points: (1, 5), (2, 8), and (3, 11)?

Answer

To find the slope for the line that has these points we will use the slope formula with two of the points.

In our case and

Now we can use the slope formula:

$m=\frac{y_2-y_1}{x_2-x_1}=\frac{8-5}{2-1}=\frac{3}{1}=3$

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Question

What is the slope of the function:

Answer

For this question we need to get the function into slope intercept form first which is

where the m equals our slope.

In our case we need to do algebraic opperations to get it into the desired form

$y=\frac{8x+16}{2}$

Therefore our slope is 4

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Question

Find the slope of the following equation:

Answer

To find the slope for a given equation, it needs to first be put into the "y=mx+b" format. Then our slope is the number in front of the x, or the "m". For this equation this looks as follows:

First subtract 2x from both sides:

That gives us the following:

Divide all three terms by three to get "y" by itself:

$\frac{3y}{3}=\frac{-2x}{3}+\frac{6}{3}\rightarrow y=\frac{-2x}{3}+2$

This means our "m" is -2/3

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Question

Find the slope of the following equation:

Answer

To find the slope for a given equation, it needs to first be put into the "y=mx+b" format. Then our slope is the number in front of the x, or the "m". For this equation this looks as follows:

First add x to both sides:

That gives us the following:

Divide all three terms by four to get "y" by itself:

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

This means our "m" is 1/4

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