Question 478·Hard·Math

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The function $g$ is defined by $g(x)=a b^{x} + c$ , where $a$ , $b$ , and $c$ are constants with $b > 0$ and $b \neq 1$ . Selected values of $g$ are given in the table.

$x$	$g(x)$
$-1$	$10$
$1$	$40$
$3$	$160$

What is the value of $g(0)$ ?

For a shifted exponential like $g(x)=ab^x+c$ , the $+c$ adds the same amount to every output. Use one table row to express one constant through the others, fit the remaining constants with a Desmos regression, then evaluate the requested input. Don't assume the raw outputs have a constant ratio: the exponential pattern applies after subtracting the shift $c$ .

Hints

Hint 1
In a shifted exponential, the added constant changes every output, so dividing one table output by another doesn't necessarily reveal the base. How can you fit all three rows without guessing the value of that constant?
Hint 2
A Desmos regression finds unknown constants from input-output pairs. Put the inputs in $x_1$ and the outputs in $y_1$ , then fit the shifted rule, including the $+c$ .
Hint 3
Function notation $g(0)$ means put $0$ in for $x$ . Once you've found the constants, what does $b^0$ become, and which constants remain?

Step-by-step

Approach 1: Fit the full function in Desmos
Step 1
Use one row to express $a$
The table says $g(-1)=10$ . Put $-1$ into the given function: $10=\frac{a}{b}+c$ . Subtract $c$ : $10-c=\frac{a}{b}$ . Multiply by $b$ : $a=(10-c)b$ . Substitute this into the rule and combine $b\cdot b^x=b^{x+1}$ : $g(x)=(10-c)b^{x+1}+c$ . The added shift remains in the rule; dropping it would change what the regression fits.
Step 2
Find the remaining constants from the table
Enter the three table rows in Desmos as $x_1,y_1$ pairs. Then type $y_1\sim(10-c)b^{x_1+1}+c\{b>0\}$ . The $\sim$ makes this a regression: Desmos finds constants that fit the rows. The restriction keeps the base positive, as required. Under PARAMETERS, Desmos shows $b=2$ and $c=0$ .
Step 3
Evaluate the requested input
Type $g(x)=(10-c)b^{x+1}+c$ , then $g(0)$ . Desmos prints $20$ . Since $b^0=1$ , the output at input $0$ is $a+c$ , not an output copied from a neighboring table row. So $g(0)=20$ . Choice B.

Approach 2: Use equal input steps and the shift
Step 1
Relate the shifted outputs
The inputs $-1,1,3$ are equally spaced. After subtracting $c$ , their outputs are $a/b$ , $ab$ , and $ab^3$ . The middle expression squared equals the product of the other two, since $(ab)^2=(a/b)(ab^3)$ . So the shifted outputs satisfy $\bigl(g(1)-c\bigr)^2=\bigl(g(-1)-c\bigr)\bigl(g(3)-c\bigr)$ .
Step 2
Find the shift
Put the table outputs into that relation: $(40-c)^2=(10-c)(160-c)$ . Expand both sides: $1600-80c+c^2=1600-170c+c^2$ . Cancel $1600$ and $c^2$ : $-80c=-170c$ . Add $170c$ : $90c=0$ . Divide by $90$ : $c=0$ .
Step 3
Use the outputs around zero
The target input $0$ is halfway between $-1$ and $1$ . Apply the same middle-output rule with $c=0$ : $g(0)^2=g(-1)g(1)$ . Substitute the table values: $g(0)^2=10\cdot40$ . Multiply: $g(0)^2=400$ .
Step 4
Choose the positive value
At $x=-1$ , $g(-1)=a/b=10$ . The positive base $b>0$ makes $a>0$ , and $g(0)=a$ because $c=0$ . Take the positive square root: $g(0)=20$ . So the function's value at input $0$ is $20$ . Choice B.

Question 478·Hard·Math

Strategy

Hints

Step-by-step

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Strategy

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Step-by-step

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