Solving Functions - SAT Math

Card 0 of 312

Question

If , what must be?

Answer

Replace the value of negative two with the x-variable.

There is no need to use the FOIL method to expand the binomial.

The answer is:

Compare your answer with the correct one above

Question

Let . What is the value of $f(\frac{7}{8})$ ?

Answer

Substitute the fraction as .

$f(\frac{7}{8}) = -3(\frac{7}{8})+9$

Multiply the whole number with the numerator.

$\frac{-21}{8}+9$

Convert the expression so that both terms have similar denominators.

$-\frac{21}{8}+\frac{9(8)}{1(8)} = -\frac{21}{8}+\frac{72}{8}$

The answer is: $\frac{51}{8}$

Compare your answer with the correct one above

Question

If , what must be?

Answer

A function of x equals five. This can be translated to:

This means that every point on the x-axis has a y value of five.

Therefore, .

The answer is:

Compare your answer with the correct one above

Question

A baseball is thrown straight up with an initial speed of 60 feet per second by a man standing on the roof of a 100-foot high building. The height of the baseball in feet as a function of time in seconds is modeled by the function

$h(t) = -16t^{2}+60t + 100$

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

Answer

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

$-16t^{2}+60t + 100 = 0$

This can be done using the quadratic formula:

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

Set :

$t = \frac{-60 \pm \sqrt{60^{2}-4(-16)100}}{2(-16)}$

$t = \frac{-60 \pm \sqrt{3,600- (- 6,400)}}{-32}$

$t = \frac{-60 \pm \sqrt{3,600+6,400}}{-32}$

$t = \frac{-60 \pm \sqrt{10,000}}{-32}$

$t = \frac{-60 \pm 100}{-32}$

One possible solution:

$t = \frac{-60 + 100}{-32} = \frac{40}{-32} = -1.25$

We throw this out, since time must be positive.

The other:

$t = \frac{-60 - 100}{-32} = \frac{-160}{-32} = 5$

This solution, we keep. The baseball hits the ground in 5 seconds.

Compare your answer with the correct one above

Question

If $f(x) = 2\sqrt{x}-\sqrt{x-3}$ , what is the value of ?

Answer

Substitute the value of negative three as .

$f(-3) = 2\sqrt{-3}-\sqrt{(-3)-3}$

$2\sqrt{-3}-\sqrt{-6} = 2\sqrt{-1}\cdot \sqrt{3} - \sqrt{-1}\cdot \sqrt{6}$

The terms will be imaginary. We can factor out an $i = \sqrt{-1}$ out of the right side. Replace them with .

The answer is: $2i\sqrt3 -i\sqrt6$

Compare your answer with the correct one above

Question

What is the horizontal asymptote of the graph of the equation $y = -2e^{x-3} -5$ ?

Answer

The asymptote of this equation can be found by observing that $e^{x-3} > 0$ regardless of . We are thus solving for the value of as $e^{x-3}$ approaches zero.

$-2e^{x-3} < -2\cdot 0$

$-2e^{x-3} < 0$

$-2e^{x-3} -5 < 0-5$

$-2e^{x-3} -5 < -5$

So the value that cannot exceed is , and the line is the asymptote.

Compare your answer with the correct one above

Question

Solve the equation for .

$\small 9^x=3^6$

Answer

Begin by recognizing that both sides of the equation have a root term of .

$\small 9^x=3^6$

Using the power rule, we can set the exponents equal to each other.

$3^{(2*x)}=3^6$

$\small 2x=6$

$\small x=3$

Compare your answer with the correct one above

Question

Solve the equation for .

$\small 3^{2x}=81$

Answer

Begin by recognizing that both sides of the equation have the same root term, .

$\small 3^{2x}=81$

$\small 3^{2x}=9^2$

$3^{2x}=(3^2)^2$

We can use the power rule to combine exponents.

$3^{2x}=3^4$

Set the exponents equal to each other.

$\small x=2$

Compare your answer with the correct one above

Question

What is/are the asymptote(s) of the graph of the function

$f(x) = 5e^{x-2}-2$ ?

Answer

An exponential equation of the form $f(x) = ae^{x-h} +k$ has only one asymptote - a horizontal one at . In the given function, , so its one and only asymptote is .

Compare your answer with the correct one above

Question

Find the vertical asymptote of the equation.

$y=\frac{x^2-9}{x^2-16}$

Answer

To find the vertical asymptotes, we set the denominator of the function equal to zero and solve.

$\sqrt{16}=\sqrt{x^2}$

Compare your answer with the correct one above

Question

In 2009, the population of fish in a pond was 1,034. In 2013, it was 1,711.

Write an exponential growth function of the form $y=ab^{x}}$ that could be used to model , the population of fish, in terms of , the number of years since 2009.

Answer

Solve for the values of a and b:

In 2009, and (zero years since 2009). Plug this into the exponential equation form:

$1034=ab^{(0)}$ . Solve for to get .

In 2013, and . Therefore,

or . Solve for to get

$b=\sqrt[4]{\frac{1711}{1034}}\approx1.1342$ .

Then the exponential growth function is

$y=1034(1.1342)^{x}$ .

Compare your answer with the correct one above

Question

Determine whether each function represents exponential decay or growth.

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

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

Answer

a)

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

b)

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

Compare your answer with the correct one above

Question

$4^{5x}=8^{4x-1}$

Solve for .

Answer

8 and 4 are both powers of 2.

$4^{5x}=8^{4x-1}$

$2^{10x}=2^{12x-3}$

Compare your answer with the correct one above

Question

Give the -intercept of the graph of the equation $f(x) = 2 ^{x- 7} + 4$ .

Answer

Set and solve for

$f(x) = 2 ^{x- 7} + 4$

$2 ^{x- 7} + 4 = 0$

$2 ^{x- 7} = -4$

We need not work further. It is impossible to raise a positive number 2 to any real power to obtain a negative number. Therefore, the equation has no solution, and the graph of has no -intercept.

Compare your answer with the correct one above

Question

What is/are the asymptote(s) of the graph of the function $g(x) = 0.4 ^{x - 2} + 0.7$ ?

Answer

An exponential function of the form

$g(x) = a^{x- h}+k$

has as its one and only asymptote the horizontal line .

Since we define as

$g(x) = 0.4 ^{x - 2} + 0.7$ ,

then ,

and the only asymptote is the line of the equation .

Compare your answer with the correct one above

Question

Match each function with its graph.

1.

2.

3.

a.

b.

c.

Answer

For , our base is greater than so we have exponential growth, meaning the function is increasing. Also, when , we know that since . The only graph that fits these conditions is .

For , we have exponential growth again but when , . This is shown on graph .

For , we have exponential decay so the graph must be decreasing. Also, when , . This is shown on graph .

Compare your answer with the correct one above

Question

What is the -intercept of $y = 4(2^{x})$ ?

Answer

The -intercept of a graph is the point on the graph where the -value is .

Thus, to find the -intercept, substitute and solve for .

Thus, we get:

$y = 4(2^{0}) = 4(1)=4$

Compare your answer with the correct one above

Question

In 2010, the population of trout in a lake was 416. It has increased to 521 in 2015.

Write an exponential function of the form that could be used to model the fish population of the lake. Write the function in terms of , the number of years since 2010.

Answer

We need to determine the constants and . Since in 2010 (when ), then and

To get , we find that when , . Then $b^5=\frac{521}{416}$ .

Using a calculator, ${\left(\frac{521}{416}\right)}^{\frac{1}{5}}=1.046$ , so $b\approx 1.046$ .

Then our model equation for the fish population is

Compare your answer with the correct one above

Question

An exponential funtion is graphed on the figure below to model some data that shows exponential decay. At , is at half of its initial value (value when ). Find the exponential equation of the form $y=Ae^{kt}$ that fits the data in the graph, i.e. find the constants and .

Answer

To determine the constant , we look at the graph to find the initial value of , (when ) and find it to be . We can then plug this into our equation $y=Ae^{kt}$ and we get $1.2=Ae^{k0}$ . Since $e^{k0}=e^0=1$ , we find that .

To find , we use the fact that when , is one half of the initial value . Plugging this into our equation with now known gives us $\frac{1.2}{2}=0.6=1.2e^{k(0.3648)}$ . To solve for , we make use the fact that the natural log is the inverse function of , so that

$\ln(e^{k(0.3648)})=k(0.3648)$ .

We can write our equation as $\frac{0.6}{1.2}=e^{k(0.3648)}$ and take the natural log of both sides to get:

$\ln(\frac{0.6}{1.2})=\ln(e^{k(0.3648)})$ or .

Then $k\approx -1.9$ .

Our model equation is $y=1.20e^{-1.9t}$ .

Compare your answer with the correct one above

Question

Consider the exponential function . Determine if there are any asymptotes and where they lie on the graph.

Answer

For positive values, increases exponentially in the direction and goes to positive infinity, so there is no asymptote on the positive -axis. For negative values, as decreases, the term becomes closer and closer to zero so approaches as we move along the negative axis. As the graph below shows, this is forms a horizontal asymptote.

Compare your answer with the correct one above

Tap the card to reveal the answer