Answer:
[tex]\displaystyle a=4, b= \frac{25}{4}, \text{ and } k = \frac{125}{2}[/tex]
Step-by-step explanation:
Note that the graph passes through the points: (0, 4), (1, 25), and (1.5, k).
The standard exponential function has the form:
[tex]\displaystyle y = ab^x[/tex]
The point (0, 4) tells us that y = 4 when x = 0. Therefore:
[tex](4) = a(b)^0[/tex]
Since anything raised to zero is one:
[tex]a=4[/tex]
Hence, our function is now:
[tex]y = 4(b)^x[/tex]
The point (1, 25) tells us that y = 25 when x = 1. By substituting:
[tex](25) = 4(b)^{(1)}[/tex]
Solve for b:
[tex]\displaystyle b = \frac{25}{4}[/tex]
Thus, our completed function is:
[tex]\displaystyle y = 4\left(\frac{25}{4}\right)^x[/tex]
To find k, simply substitute 1.5 for x. This yields:
[tex]\displaystyle y = k = 4\left(\frac{25}{4}\right)^{(1.5)}[/tex]
And evaluate. Hence:
[tex]\displaystyle \begin{aligned} k &= 4\left(\frac{25}{4}\right)^{3/2} \\ \\ &= 4\left(\left(\frac{25}{4}\right)^{1/2}\right)^3 \\ \\ &= 4\left(\frac{5}{2}\right)^3 \\ \\ &= 4\left(\frac{125}{8}\right) \\ \\ &= \frac{125}{2}\end{aligned}[/tex]
In conclusion:
[tex]\displaystyle a=4, b= \frac{25}{4}, \text{ and } k = \frac{125}{2}[/tex]
