Question 37·SAT Desmos Regression Practice·Advanced Math

Each function value gives an input and its output. So $f(0)=7$, $f(1)=13$, and $f(3)=61$ give the three points $(0,7)$, $(1,13)$, and $(3,61)$ for a table.

Enter those points in a table, then type $y_1\sim ab^{x_1}+k\{b>0\}$. The $\sim$ tells Desmos to fit the constants, and the braces enforce the positive base the problem requires. Under PARAMETERS, Desmos reports $b\approx2.37228$.

Type $\frac{-1+\sqrt{33}}{2}$. Desmos prints approximately $2.37228$, matching the fitted $b$; the other radical is negative, while $3$ and $8$ do not match. So the value of $b$ is $\frac{-1+\sqrt{33}}{2}$. Choice A.

In $f(x)=ab^x+k$, the same $k$ is added to every output. Subtracting outputs cancels that constant. Subtract $f(0)$ from each later output:

Both differences contain $a$, so divide the larger change by the smaller one: $\frac{a(b^3-1)}{a(b-1)}=\frac{54}{6}$. Simplify the right side:

Factor the difference of cubes: $b^3-1=(b-1)(b^2+b+1)$. Substitute it into the ratio: $\frac{a(b-1)(b^2+b+1)}{a(b-1)}=9$. The denominator is $6$, so it is not zero. Cancel $a(b-1)$: $b^2+b+1=9$. The ratio is not $b^2$: the change from input $0$ to input $3$ includes three one-step changes.

Subtract $9$ from both sides to make one side zero: $b^2+b-8=0$. This is a quadratic equation, with a $b^2$ term, so it can have two solutions.

Apply the quadratic formula to $b^2+b-8=0$, using coefficients $1$, $1$, and $-8$: $b=\frac{-1\pm\sqrt{1^2-4(1)(-8)}}{2(1)}$. Simplify under the square root: $b=\frac{-1\pm\sqrt{33}}{2}$. Keep both signs until you check the condition on $b$.

Since $\sqrt{33}>1$, the root with the minus sign is negative. The problem requires $b>0$, so the value of the base is $\frac{-1+\sqrt{33}}{2}$. Choice A.