SAT Math - Solving Functions

Find the point at which these two lines intersect:

$\small 2x+5y=8$

$\small \small \small -2x+2y=-1$

$\left(\frac{3}{2},1\right)$

$\left(-\frac{3}{2},1\right)$

$\left(1,\frac{3}{2}\right)$

$\left(1,-\frac{3}{2}\right)$

Explanation

We are looking for a point, $\small (x,y)$ , where these two lines intersect. While there are many ways to solve for $\small x$ and $\small y$ given two equations, the simplest way I see is to use the elimination method since by adding the two equations together, we can eliminate the $\small x$ variable.

Dividing both sides by 7, we isolate y.

$\small y=1$

Now, we can plug y back into either equation and solve for x.

$\small 2x+(5)(1)=8$

$\small 2x+5=8$

Next, we can solve for x.

$\small 2x=3$

$\small x=(3/2)$

Therefore, the point where these two lines intersect is $\left(\frac{3}{2}, 1\right)$ .

Find the point at which these two lines intersect:

$\small 2x+5y=8$

$\small \small \small -2x+2y=-1$

$\left(\frac{3}{2},1\right)$

$\left(-\frac{3}{2},1\right)$

$\left(1,\frac{3}{2}\right)$

$\left(1,-\frac{3}{2}\right)$

Explanation

Dividing both sides by 7, we isolate y.

$\small y=1$

Now, we can plug y back into either equation and solve for x.

$\small 2x+(5)(1)=8$

$\small 2x+5=8$

Next, we can solve for x.

$\small 2x=3$

$\small x=(3/2)$

Therefore, the point where these two lines intersect is $\left(\frac{3}{2}, 1\right)$ .

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:

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

$\small x=2\or-6$

Determine whether each function represents exponential decay or growth.

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

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 .

Define functions and as follows:

$f(x) = \left{\begin{matrix} 2x+ 1 &x< 0 \ 3x & x \ge 0 \end{matrix}\right.$

$g(x) = \left{\begin{matrix} x+ 9 &x< 0 \ x- 2 & x \ge 0 \end{matrix}\right.$

Evaluate $(f \circ g) \left (-1.7\right )$ .

Undefined

Explanation

$(f \circ g) \left ( -1.7\right )=f\left ( g \left ( -1.7\right ) \right )$

First we evaluate . Since , we use the definition of for the values in the range :

Therefore,

$(f \circ g) \left ( -1.7\right )=f\left ( g \left ( -1.7\right ) \right ) = f (7.3)$

Since $7.3 \ge 0$ , we use the definition of for the range $x \ge 0$ :

$f (7.3) = 3 \cdot 7.3 = 21.9$

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.

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: $\frac{8}{3}$

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 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 .

Define functions and as follows:

$f(x) = \left{\begin{matrix} 2x+ 1 &x< 0 \ 3x & x \ge 0 \end{matrix}\right.$

$g(x) = \left{\begin{matrix} x+ 9 &x< 0 \ x- 2 & x \ge 0 \end{matrix}\right.$

Evaluate $(f \circ g) \left (-1.7\right )$ .

Undefined

Explanation

$(f \circ g) \left ( -1.7\right )=f\left ( g \left ( -1.7\right ) \right )$

First we evaluate . Since , we use the definition of for the values in the range :

Therefore,

$(f \circ g) \left ( -1.7\right )=f\left ( g \left ( -1.7\right ) \right ) = f (7.3)$

Since $7.3 \ge 0$ , we use the definition of for the range $x \ge 0$ :

$f (7.3) = 3 \cdot 7.3 = 21.9$

Solving Functions

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