The function is defined by where and are nonzero integers and is a real constant. The functions and are equivalent to . Which of the following equations displays the -coordinate of the -intercept of the graph of in the -plane as a constant or coefficient?

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SAT Nonlinear Functions Question 174 | Free SAT Prep | Aniko

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Question 174·Hard·Nonlinear Functions

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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\bigl(3^{x}-1\bigr)+\bigl(r+t-t\cdot 3^{s}\bigr) \\[0.7em] \text{II. } h(x)=t\cdot 3^{s}\bigl(3^{x-s}-1\bigr)+r

Which of the following equations displays the $y$ -coordinate of the $y$ -intercept of the graph of $y=f(x)$ in the $xy$ -plane as a constant or coefficient?

For questions about intercepts in rewritten function forms, first express the y-intercept symbolically by computing $f(0)$ from the original definition. Then examine each proposed equivalent form and substitute $x=0$ mentally: check whether a whole term becomes $0$ , leaving a single visible constant or coefficient that matches $f(0)$ . This lets you decide quickly which forms truly display the y-intercept without doing full expansions or heavy algebra.

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Start with the definition of y-intercept

For any function $y = f(x)$ , what $x$ -value do you use to find the y-coordinate of the y-intercept? Write $f(\text{that value})$ for this function.

Compute $f(0)$ from the original formula

Substitute $x=0$ into $f(x) = t(3^x - 3^s) + r$ and simplify as much as possible. This expression is the y-coordinate of the y-intercept.

Relate $f(0)$ to the forms of $g$ and $h$

After you find $f(0)$ , look at $g(x)$ and $h(x)$ . In which form does substituting $x=0$ make one whole term become $0$ , leaving a single constant equal to $f(0)$ ?

Focus on which constant really equals the y-intercept

Be careful: just because there is a "+ r" at the end of an expression does not mean the y-intercept is $r$ . Check whether any other part of the expression is nonzero when $x=0$ .

Desmos Guide

Set up parameters and the original function

In Desmos, create sliders t, s, and r. Make t and s integer sliders (and make sure s is not zero), and r any real slider. Then enter f(x) = t(3^x - 3^s) + r.

Graph the equivalent forms

Enter g(x) = t(3^x - 1) + (r + t - t*3^s) and h(x) = t*3^s(3^(x - s) - 1) + r. You should see that all three graphs coincide for the same values of t, s, and r (they are equivalent functions).

Compare the y-intercepts with the constants in each form

In Desmos, evaluate f(0), g(0), and h(0) by typing them directly. Note the numeric value of the y-intercept. Then compare that number to the constant terms you see in g(x) and h(x) (for g(x), the standalone constant outside the parentheses; for h(x), the + r part). Identify which equation has a single constant that equals the y-intercept value from f(0).

Step-by-step Explanation

Find the y-coordinate of the y-intercept for $f$

The y-intercept occurs where the graph crosses the $y$ -axis, which is always at $x=0$ .

So we find the y-coordinate by computing $f(0)$ :

f(0) = t(3^{0} - 3^{s}) + r

Since $3^0 = 1$ ,

f(0) = t(1 - 3^{s}) + r = r + t - t\cdot 3^{s}.

This value $r + t - t\cdot 3^{s}$ is the y-coordinate of the y-intercept of $y = f(x)$ .

Understand what it means to "display" the y-intercept

The question asks which equation displays the y-coordinate of the y-intercept as a constant or coefficient.

This means: in that version of the function, you can point to a single number (a constant term or coefficient) in the equation that is equal to $f(0)$ , without having to plug in $x=0$ and simplify.

A common structure that does this is something like

y = A(3^x - 1) + B,

because when $x=0$ , $3^0 - 1 = 0$ , so the first term disappears and the y-intercept is just $B$ .

Check equation I: $g(x)$

Equation I is

g(x) = t(3^{x} - 1) + (r + t - t\cdot 3^{s}).

Plug in $x=0$ :

g(0) = t(3^{0} - 1) + (r + t - t\cdot 3^{s}) = t(1 - 1) + (r + t - t\cdot 3^{s}).

Since $1 - 1 = 0$ ,

g(0) = 0 + (r + t - t\cdot 3^{s}) = r + t - t\cdot 3^{s}.

From Step 1, this equals $f(0)$ , the y-coordinate of the y-intercept.

In this form, the term $t(3^{x} - 1)$ becomes $0$ at $x=0$ , so the constant $(r + t - t\cdot 3^{s})$ is exactly the y-intercept. Therefore equation I does display the y-intercept as a constant.

Check equation II: $h(x)$ and conclude

Equation II is

h(x) = t\cdot 3^{s}(3^{x-s} - 1) + r.

Plug in $x=0$ :

h(0) = t\cdot 3^{s}(3^{0-s} - 1) + r = t\cdot 3^{s}(3^{-s} - 1) + r.

Because $s$ is a nonzero integer, $3^{-s} \ne 1$ , so $(3^{-s} - 1) \ne 0$ . That means the first term does not vanish at $x=0$ , and the y-intercept is not just the constant $r$ .

We already know the y-intercept is $r + t - t\cdot 3^{s}$ , which is different from $r$ alone, so equation II does not show the y-intercept as a single constant or coefficient that you can read off directly.

So only equation I has a constant term that equals the y-coordinate of the y-intercept.

Correct answer: I only.

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