Answer:
The fourth choice.
Step-by-step explanation:
An exponential function passes through (3, 5) and (4, 10).
And we want to determine which other point is also on the graph.
First, we can find the exponential function. A standard exponential function is in the form:
[tex]y=a(b)^x[/tex]
The point (3, 5) tells us that y = 5 when x = 3. Thus:
[tex]5=a(b)^3[/tex]
The point (4, 10) tells us that y = 10 when x = 4. Thus:
[tex]10=a(b)^4[/tex]
In the first equation, we can divide both sides by a:
[tex]\displaystyle \frac{5}{a}=b^3[/tex]
And we can rewrite the second equation:
[tex]\displaystyle 2(5)=a(b^4)\Rightarrow 2\Big(\frac{5}{a}\Big)=b^4[/tex]
Substitute:
[tex]2b^3=b^4[/tex]
Divide:
[tex]b=2[/tex]
Using the first equation again, substitute:
[tex]5=a(2)^3[/tex]
Simplify and solve:
[tex]5=a(8)\Rightarrow \displaystyle a=\frac{5}{8}[/tex]
So, our exponential function is:
[tex]\displaystyle y=\frac{5}{8}(2)^x[/tex]
Next, we can simply try each point and see which point is correct.
Testing the first point (remember that (2,0) means that y = 0 when x = 2):
[tex]\displaystyle 0\stackrel{?}{=}\frac{5}{8}(2)^0=\frac{5}{8}(1)=\frac{5}{8}\neq 0[/tex]
Testing the second point:
[tex]\displaystyle 1\stackrel{?}{=}\frac{5}{8}(2)^2=\frac{5}{8}(4)=\frac{5}{2}\neq 1[/tex]
The third point:
[tex]\displaystyle 15\stackrel{?}{=}\frac{5}{8}(2)^5=\frac{5}{8}(32)=5(4)=20\neq 15[/tex]
And the fourth point:
[tex]\displaystyle 20\stackrel{?}{=}\frac{5}{8}(2)^5=\frac{5}{8}(32)=5(4)=20\stackrel{\checkmark}{=}20[/tex]
Therefore, D is the correct choice.
