1. The problem is to find the values of $a$, $b$, and $k$ in the exponential function $f(x) = a b^x + k$ given the table of values: $$\begin{array}{c|ccccc} x & 0 & 1 & 2 & 3 & 4 \\ \hline f(x) & 6 & 12 & 36 & 132 & 516 \end{array}$$ 2. We start by using the value at $x=0$: $$f(0) = a b^0 + k = a \cdot 1 + k = a + k = 6$$ This gives the equation: $$k = 6 - a$$ 3. Next, use the value at $x=1$: $$f(1) = a b^1 + k = a b + k = 12$$ Substitute $k$ from step 2: $$12 = a b + (6 - a) = a b + 6 - a$$ Rearranged: $$12 - 6 = a b - a$$ $$6 = a (b - 1)$$ 4. The user suggests $b = 4$ (from the arrows indicating multiplication by 4 between function values), so substitute $b=4$: $$6 = a (4 - 1) = 3a$$ Solve for $a$: $$a = \frac{6}{3} = 2$$ 5. Substitute $a=2$ back into $k = 6 - a$: $$k = 6 - 2 = 4$$ 6. Check if this fits the next value $f(2)$: $$f(2) = a b^2 + k = 2 \cdot 4^2 + 4 = 2 \cdot 16 + 4 = 32 + 4 = 36$$ This matches the table value. 7. Explanation why $a$ is not 6: - At $x=0$, $f(0) = a + k = 6$, so if $a$ were 6, then $k$ would be 0. - But using $x=1$, $f(1) = a b + k = 12$, substituting $a=6$ and $k=0$ gives $6 \cdot 4 = 24$, which does not equal 12. - Therefore, $a$ cannot be 6; it must be 2 to satisfy both equations. Final answer: $$a = 2, \quad b = 4, \quad k = 4$$