Question

f(x) = 5(3)^{x+3}
for the given function f, which of the following equivalent forms shows the y-coordinate of the y-intercept of the graph of y = f(x) in the xy-plane as a constant or coefficient?
a) f(x) = \frac{5}{3} (3)^{x + 4}
b) f(x) = 15(3)^{x + 2}
c) f(x) = 45(3)^{x + 1}
d) f(x) = 135(3)^{x}

Explanation:

Step1: Recall the y - intercept formula

The y - intercept of a function \(y = f(x)\) occurs when \(x = 0\). So we need to find the value of \(f(0)\) for the function \(f(x)=5(3)^{x + 3}\) and then check which of the given options, when \(x = 0\), has the coefficient or constant equal to \(f(0)\).

First, let's simplify the original function using the exponent rule \(a^{m + n}=a^{m}\times a^{n}\). So \(f(x)=5\times3^{x}\times3^{3}\). Since \(3^{3}=27\), then \(f(x)=5\times27\times3^{x}=135\times3^{x}\).

Step2: Analyze the y - intercept

To find the y - intercept, we set \(x = 0\) in the original function \(f(x)=5(3)^{x + 3}\). Then \(f(0)=5(3)^{0 + 3}=5\times3^{3}=5\times27 = 135\).

Now let's check each option:

• Option A: When \(x = 0\), \(f(0)=\frac{5}{3}(3)^{0 + 4}=\frac{5}{3}\times3^{4}=\frac{5}{3}\times81 = 135\)? Wait, no, \(\frac{5}{3}\times81=5\times27 = 135\)? Wait, \(\frac{5}{3}\times3^{4}=\frac{5}{3}\times81 = 135\), but the form is \(\frac{5}{3}(3)^{x + 4}\), when we simplify the original function, we can also use exponent rules. Wait, maybe a better way is to rewrite the original function.

Original function: \(f(x)=5(3)^{x + 3}=5\times3^{x}\times3^{3}=5\times27\times3^{x}=135\times3^{x}\), which is option D. When \(x = 0\), \(f(0)=135\times3^{0}=135\times1 = 135\), and in the form \(f(x)=135(3)^{x}\), the coefficient 135 is the y - intercept (since when \(x = 0\), \(y = 135\)).

Let's check other options:

• Option B: \(f(x)=15(3)^{x + 2}=15\times3^{x}\times3^{2}=15\times9\times3^{x}=135\times3^{x}\)? Wait, \(15\times9 = 135\), but the form is \(15(3)^{x + 2}\). When \(x = 0\), \(f(0)=15(3)^{2}=15\times9 = 135\)? Wait, no, \(15\times9 = 135\), but the form is different. Wait, no, let's simplify \(15(3)^{x+2}=15\times3^{x}\times3^{2}=15\times9\times3^{x}=135\times3^{x}\), but the question is about the form that shows the y - intercept as a coefficient or constant. The y - intercept is when \(x = 0\), so in the form \(f(x)=a(3)^{x}\), when \(x = 0\), \(y=a\). So we need the form where the exponent of 3 is just \(x\), so that when \(x = 0\), the term with 3 becomes 1, and the coefficient \(a\) is the y - intercept.

In option D, \(f(x)=135(3)^{x}\), when \(x = 0\), \(y = 135\times1=135\), which is the y - intercept. Let's verify with the original function: \(f(0)=5(3)^{0 + 3}=5\times27 = 135\), which matches.

Option A: \(f(x)=\frac{5}{3}(3)^{x + 4}=\frac{5}{3}\times3^{x}\times3^{4}=\frac{5}{3}\times81\times3^{x}=135\times3^{x}\), but the form is \(\frac{5}{3}(3)^{x + 4}\), which is equivalent, but the question is which form shows the y - intercept as a coefficient or constant. In option D, the exponent of 3 is \(x\), so when \(x = 0\), the 3 term is 1, and the coefficient is the y - intercept. In option D, the form is \(135(3)^{x}\), so the coefficient 135 is the y - intercept.

Answer:

D. \(f(x) = 135(3)^{x}\)