Question

the function below has the form $f(x) = a \cdot b^x$. which of the following functions is shown on the graph? $\circ f(x) = 3 \cdot 4^x$ $\circ f(x) = -3 \cdot \left(\frac{1}{4}\
ight)^x$ $\circ f(x) = 3 \cdot \left(\frac{1}{4}\
ight)^x$ $\circ f(x) = -\frac{1}{3} \cdot \left(\frac{1}{4}\
ight)^x$ $\circ f(x) = -3 \cdot 4^x$ $\circ f(x) = \frac{1}{3} \cdot \left(\frac{1}{4}\
ight)^x$ $\circ f(x) = \frac{1}{3} \cdot 4^x$ $\circ f(x) = -\frac{1}{3} \cdot 4^x$

Explanation:

Step1: Analyze the sign of \(a\)

The graph is decreasing (since as \(x\) increases, \(y\) decreases rapidly) and below the \(x\)-axis for large \(x\), so \(a\) should be negative. Eliminate options with \(a>0\) (first, third, sixth, seventh).

Step2: Analyze the base \(b\)

The graph has a rapid decrease, indicating exponential decay with a large base (or growth with negative exponent, but here the shape suggests \(b > 1\) and \(a<0\) for rapid decrease). The base \(\frac{1}{4}\) would give slow decay, while \(4\) gives rapid change. Check the remaining options: second has \(b=\frac{1}{4}\) (slow), fourth has small \(|a|\), fifth has \(a = -3\), \(b = 4\).

Step3: Verify with a point

Take \(x = 0\): \(f(0)=a\cdot b^0=a\). The graph passes near \(y=-3\) at \(x = 0\) (since at \(x=0\), \(f(0)=a\)). For \(f(x)=-3\cdot 4^x\), \(f(0)=-3\), which matches the graph's \(y\)-intercept (near \(-3\)). The other negative \(a\) options: \(f(x)=-3\cdot(\frac{1}{4})^x\) at \(x = 0\) is \(-3\), but its decay is slow (wrong shape), \(f(x)=-\frac{1}{3}\cdot(\frac{1}{4})^x\) has \(f(0)=-\frac{1}{3}\) (too small), \(f(x)=-\frac{1}{3}\cdot 4^x\) has \(f(0)=-\frac{1}{3}\) (too small). So \(f(x)=-3\cdot 4^x\) matches.

Answer:

$f(x) = -3 \cdot 4^x$