Question 694·Hard·Math

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The function $f$ is defined by

f(x)=t\bigl(3^{x}-3^{s}\bigr)+r,

where $t$ and $s$ are nonzero integers and $r$ is a real constant. The functions $g$ and $h$ are equivalent to $f$ .

\text{I. } g(x)=t\cdot 3^x+\bigl(r-t\cdot 3^s\bigr) \\[1.4em] \text{II. } h(x)=t\bigl(3^x-1\bigr)+\bigl(r+t-t\cdot 3^s\bigr)

As $x$ decreases without bound, the graph of $y=f(x)$ approaches a particular $y$ -coordinate. For each equation, consider whether it displays that $y$ -coordinate as a single constant term that can be read directly without combining terms.

Which of the following correctly identifies the equation or equations that do so?

An exponential graph’s horizontal asymptote is the height it approaches as the input heads far left or right. Since the base here is greater than $1$ , its power approaches $0$ as $x$ heads left. Find the height that remains, then ask which form shows it as one standalone constant term. Don’t read a parenthesized constant as the answer until you’ve checked what the rest of the expression approaches.

Hints

Hint 1
For a base greater than $1$ , each move left makes $3^x$ smaller. As $x$ decreases without bound, what does $3^x$ approach, and what part of $f(x)$ remains?
Hint 2
A constant term has no $x$ in it. In $g(x)$ , the parentheses group $r-t\cdot3^s$ into one term. Does any other part of $g(x)$ contribute to the height the graph approaches?
Hint 3
In $h(x)$ , don’t assume the part containing $3^x$ disappears entirely. When $3^x$ approaches $0$ , what does $3^x-1$ approach? Check that before reading the constant in parentheses.

Step-by-step

Find the left-end height, then inspect each form
Step 1
Determine what happens to the power
No Desmos needed. The left-end behavior follows from the exponential rule without choosing values for the constants.

Because the base $3$ is greater than $1$ , each move left divides its power by $3$ . So $3^x$ approaches $0$ as $x$ decreases without bound.
Step 2
Find the height the graph approaches
Replace $3^x$ in $f(x)$ by the $0$ it approaches:
$\lim_{x\to-\infty}f(x)=t(0-3^s)+r.$
Simplify the product:
$\lim_{x\to-\infty}f(x)=-t3^s+r.$
Reorder the sum to identify the left-end height:
$\lim_{x\to-\infty}f(x)=r-t3^s.$
Step 3
Inspect the constant term in I
In $g(x)=t\cdot3^x+(r-t\cdot3^s)$ , the first term approaches $0$ . The parentheses make $r-t\cdot3^s$ one grouped constant term, so you can read the height directly from this form without combining it with another term.
Step 4
Check the part outside the constant in II
In $h(x)$ , the first group does not approach $0$ : the $-1$ remains even when $3^x$ approaches $0$ . Replace $3^x$ by $0$ :
$\lim_{x\to-\infty}h(x)=t(0-1)+(r+t-t3^s).$
Simplify the first group:
$\lim_{x\to-\infty}h(x)=-t+(r+t-t3^s).$
Cancel $-t$ with $+t$ :
$\lim_{x\to-\infty}h(x)=r-t3^s.$
The constant in $h$ ’s parentheses is not the height by itself; its other group contributes $-t$ . Since $t$ is nonzero, you must combine terms to get that height. Only I displays it directly. Choice A.

Question 694·Hard·Math

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