Functions and Graphs

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SAT Math › Functions and Graphs

Questions 1 - 10

What is the domain of the following function? Please use interval notation.

$y=\left | -x\right |$

$(-\infty ,\infty )$

$(0,\infty )$

$(-\infty ,0)$

Explanation

A basic knowledge of absolute value and its functions is valuable for this problem. However, if you do not know what the typical shape of an absoluate value function looks like, one can always plug in values and plot points.

Upon doing so, we learn that the -values (domain) are not restricted on either end of the function, creating a domain of negative infinity to postive infinity.

If we plug in -100000 for , we get 100000 for .

If we plug in 100000 for , we get 100000 for .

Additionally, if we plug in any value for , we will see that we always get a real, defined value for .

**Extra Note: Due to the absolute value notation, the negative (-) next to the is not important, in that it will always be made positive by the absolute value, making this function the same as $y=\left | x\right |$ . If the negative (-) was outside of the absolute value, this would flip the function, making all corresponding -values negative. However, this knowledge is most important for range, rather than domain.

Solve: $2^{-3x+6} = \frac{1}{4}$

$\frac{8}{3}$

$\frac{1}{2}$

$\frac{5}{6}$

$-\frac{4}{3}$

$\frac{4}{3}$

Explanation

Rewrite the right side as base 2.

$\frac{1}{4} = 2^{-2}$

Replace the term into the equation.

$2^{-3x+6} = 2^{-2}$

With similar bases, we can set the exponents equal.

Subtract six from both sides.

Divide by negative three on both sides.

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

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

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?

$4.7\textrm{ sec}$

$3.1\textrm{ sec}$

$1.6\textrm{ sec}$

$7.5 \textrm{ sec}$

$6.4\textrm{ sec}$

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 :

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

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

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

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

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

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.

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

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

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

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

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

Explanation

Factor:

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

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:

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

$\small x=2\or-6$

Solve: $2^{-3x+6} = \frac{1}{4}$

$\frac{8}{3}$

$\frac{1}{2}$

$\frac{5}{6}$

$-\frac{4}{3}$

$\frac{4}{3}$

Explanation

Rewrite the right side as base 2.

$\frac{1}{4} = 2^{-2}$

Replace the term into the equation.

$2^{-3x+6} = 2^{-2}$

With similar bases, we can set the exponents equal.

Subtract six from both sides.

Divide by negative three on both sides.

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

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

Solve the following function:

$11=x^{2}-5$

and

Explanation

You must get by itself so you must add to both side which results in

$16=x^{2}$ .

You must get the square root of both side to undue the exponent.

This leaves you with .

But since you square the in the equation, the original value you plug can also be its negative value since squaring it will make it positive anyway.

This means your answer can be or .

Determine whether each function represents exponential decay or growth.

$\newline a) \newline y=(\frac{1}{2})^{x}$

$\newline b) \newline y=0.4(1.1)^x$

a) decay

b) growth

a) growth

b) growth

a) decay

b) decay

a) growth

b) decay

Explanation

This is exponential decay since the base, , is between and .

This is exponential growth since the base, , is greater than .

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

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

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

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

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

Explanation

Factor:

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

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:

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

$\small x=2\or-6$

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

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

$4.7\textrm{ sec}$

$3.1\textrm{ sec}$

$1.6\textrm{ sec}$

$7.5 \textrm{ sec}$

$6.4\textrm{ sec}$

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 :

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

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

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

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

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

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.

Consider the equation:

$y = 3x^{2} - 10x +5$

The vertex of this parabolic function would be located at:

$(\frac{5}{3}, \frac{-10}{3})$

$(\frac{-5}{3}, \frac{-10}{3})$

$(\frac{-5}{3}, 30)$

$(\frac{5}{3}, \frac{2}{25})$

$(\frac{5}{3}, \frac{27}{25})$

Explanation

For any parabola, the general equation is

$y = ax^{2} + bx +c$ , and the x-coordinate of its vertex is given by

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

For the given problem, the x-coordinate is

$x = \frac{-(-10)}{2(3)} = \frac{5}{3}$ .

To find the y-coordinate, plug $\frac{5}{3}$ into the original equation:

$y = 3(\frac{5}{3})^{2} - 10(\frac{5}{3}) +5 = \frac{-10}{3}$

Therefore the vertex is at $(\frac{5}{3}, \frac{-10}{3})$ .

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