The function is defined by the equation above for . The constants and are positive, and . Which of the following must be true? I. The graph of has a horizontal asymptote at . II. The function does not attain a maximum value on the domain . III. The function is decreasing for all .

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Super-Hard SAT Math Question 192 | Free SAT Prep | Aniko

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Question 192·200 Super-Hard SAT Math Questions·Advanced Math

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h(x)=p-q\cdot r^{x^2}

The function $h$ is defined by the equation above for $x\ge 0$ . The constants $p$ and $q$ are positive, and $0<r<1$ . Which of the following must be true?

I. The graph of $h$ has a horizontal asymptote at $y=p$ .

II. The function $h$ does not attain a maximum value on the domain $x\ge 0$ .

III. The function $h$ is decreasing for all $x\ge 0$ .

When a function includes an exponential term with $0<r<1$ , first decide whether that term increases or decreases as the input grows. Then translate that into how the whole function behaves (increasing/decreasing, approaching a constant, or being bounded). Finally, be careful to distinguish a horizontal asymptote (a value the function approaches) from an actual maximum (a value the function reaches at some input).

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Track what happens to the exponential term

Focus on $r^{x^2}$ with $0<r<1$ . What happens to this value as $x$ gets larger?

Think about an upper bound

Compare $h(x)=p-q\cdot r^{x^2}$ to $p$ . Can $h(x)$ ever equal $p$ for a finite value of $x$ ?

Separate “approaches” from “reaches”

A function can get arbitrarily close to a number without ever attaining it. Decide which situation is happening here.

Desmos Guide

Enter the function with sliders

In Desmos, enter h(x)=p-q*r^(x^2).

Desmos will create sliders for $p$ , $q$ , and $r$ .

Restrict to the domain $x\ge 0$

Enter x>=0 as a restriction by writing h(x)=p-q*r^(x^2) {x>=0}.

Test the asymptote idea visually

Set sliders to values that satisfy the conditions (for example, $p=5$ , $q=3$ , and choose $r$ between 0 and 1).

Then graph the line y=p and zoom out to the right. Notice the curve gets closer to y=p as $x$ increases.

Check “maximum reached” vs “approached”

Create a table of $x$ -values (such as $0,1,2,3,4,5$ ) and look at the corresponding $h(x)$ values.

As you increase $x$ , the values should get closer to $p$ but remain less than $p$ , supporting that the function does not attain a maximum value.

Step-by-step Explanation

Describe the behavior of $r^{x^2}$

Since $x\ge 0$ , we have $x^2\ge 0$ . With $0<r<1$ , larger exponents make $r^{\text{exponent}}$ smaller.

So as $x$ increases, $r^{x^2}$ decreases, starting at $r^0=1$ and getting closer and closer to $0$ .

Find the end behavior of $h(x)$

The function is

h(x)=p-q\cdot r^{x^2}.

As $x\to\infty$ , we have $r^{x^2}\to 0$ , so $q\cdot r^{x^2}\to 0$ and therefore

h(x)\to p.

So the graph approaches the horizontal line $y=p$ as $x$ increases.

Decide whether a maximum value is reached

For any real $x$ , $r^{x^2}>0$ , and since $q>0$ this implies $q\cdot r^{x^2}>0$ . Therefore

h(x)=p-q\cdot r^{x^2}<p

for every $x\ge 0$ .

So $p$ is an upper bound that the function approaches but never equals on the domain, which means the function does not attain a maximum value on $x\ge 0$ .

Check statement III and choose the correct option

Because $r^{x^2}$ decreases as $x$ increases, the quantity $-q\cdot r^{x^2}$ increases (it becomes less negative), so $h(x)$ increases on $x\ge 0$ rather than decreases. Thus statement III is false.

Therefore, the statements that must be true are I and II only, so the correct choice is I and II only.

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Desmos Guide

Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation