An exponential function is defined by where and . The graph of passes through the points , , and . Which choice is the value of ?

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SAT Nonlinear Functions Question 45 | Free SAT Prep | Aniko

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Question 45·Hard·Nonlinear Functions

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An exponential function is defined by

f(x)=a\cdot b^x+c

where $b>0$ and $b\ne 1$ . The graph of $f$ passes through the points $(0,5)$ , $(1,17)$ , and $(3,377)$ . Which choice is the value of $c$ ?

When an exponential function includes a vertical shift $c$ , subtracting function values at different $x$ -coordinates is a fast way to remove $c$ entirely. After that, look for a ratio that cancels the remaining scale factor $a$ , leaving an equation in only the base $b$ . Simplify expressions like $b^3-b$ by factoring, use any given restrictions (here, $b>0$ ), and then back-substitute to find the shift.

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Eliminate the shift

Try subtracting $f(0)$ from $f(1)$ and subtracting $f(1)$ from $f(3)$ . What happens to $c$ ?

Create a ratio

After you get two expressions involving $a$ and $b$ , divide one by the other so that $a$ cancels.

Factor to simplify

If you see $b^3-b$ , factor it as $b(b^2-1)$ and then as $b(b-1)(b+1)$ .

Use the condition on $b$

If solving gives two possible values of $b$ , use the condition $b>0$ to decide which one is allowed.

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Rewrite using a single variable for $c$

Let $x$ represent $c$ . Enter

b=\frac{17-x}{5-x}

(This comes from solving $(5-x)b+x=17$ for $b$ .)

Build the expression for $f(3)$

Enter

y_1=(5-x)\cdot b^3+x

This equals $f(3)$ because $a=5-x$ when $f(0)=5$ .

Graph the target value

Enter

y_2=377

Find the intersection

Find the intersection point of $y_1$ and $y_2$ . The $x$ -coordinate of that intersection is the value of $c$ .

Step-by-step Explanation

Use differences to eliminate $c$

From $f(x)=a\cdot b^x+c$ :

\begin{aligned} f(1)-f(0) &= (ab+c)-(a+c)=a(b-1) \\ f(3)-f(1) &= (ab^3+c)-(ab+c)=a(b^3-b) \end{aligned}

Using the given points:

\begin{aligned} f(1)-f(0) &= 17-5=12 \\ f(3)-f(1) &= 377-17=360 \end{aligned}

Form a ratio to solve for $b$

Divide the two difference equations:

\frac{360}{12}=\frac{a(b^3-b)}{a(b-1)}=\frac{b^3-b}{b-1}

Cancel $a$ (since $a\ne 0$ ), and factor:

\frac{b^3-b}{b-1}=\frac{b(b-1)(b+1)}{b-1}=b(b+1)

So $30=b(b+1)$ , which gives

b^2+b-30=0

Thus $(b-5)(b+6)=0$ , so $b=5$ or $b=-6$ . Since $b>0$ , $b=5$ .

Find $a$ using $f(1)-f(0)=12$

From Step 1, $f(1)-f(0)=a(b-1)=12$ .

With $b=5$ :

a(5-1)=12 \Rightarrow 4a=12 \Rightarrow a=3

Use $f(0)=5$ to find $c$

Since $f(0)=a\cdot b^0+c=a+c=5$ and $a=3$ :

3+c=5 \Rightarrow c=2

Therefore, the correct choice is $2$ .

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