Question

1. write an exponential equation in the form $y = ab^x$ whose graph passes through points $(-3, 24)$ and $(-2, 12)$

$\bigcirc$ $y = 3(0.5)^x$
$\bigcirc$ $y = 0.5(3)^x$
$\bigcirc$ $y = 3^x$
$\bigcirc$ $y = -3(24)^x$

Explanation:

Step1: Substitute the first point \((-3, 24)\) into \( y = ab^x \)

We get the equation \( 24 = ab^{-3} \), which can be rewritten as \( 24 = \frac{a}{b^3} \) (since \( b^{-n}=\frac{1}{b^n} \)).

Step2: Substitute the second point \((-2, 12)\) into \( y = ab^x \)

We get the equation \( 12 = ab^{-2} \), which can be rewritten as \( 12 = \frac{a}{b^2} \) (since \( b^{-n}=\frac{1}{b^n} \)).

Step3: Divide the two equations to eliminate \( a \)

Divide the equation from Step 2 by the equation from Step 1: \(\frac{12}{24}=\frac{\frac{a}{b^2}}{\frac{a}{b^3}}\).
Simplify the left side: \(\frac{12}{24}=\frac{1}{2}\).
Simplify the right side: \(\frac{\frac{a}{b^2}}{\frac{a}{b^3}}=\frac{a}{b^2}\times\frac{b^3}{a}=b\) (the \( a \) terms cancel out).
So we find that \( b=\frac{1}{2}=0.5 \).

Step4: Substitute \( b = 0.5 \) back into one of the equations to find \( a \)

Let's use the equation from Step 2: \( 12=\frac{a}{(0.5)^2} \).
First, calculate \((0.5)^2 = 0.25\). Then the equation becomes \( 12=\frac{a}{0.25} \).
Multiply both sides by \( 0.25 \) to solve for \( a \): \( a = 12\times0.25 = 3 \).

Step5: Write the exponential equation

Now that we have \( a = 3 \) and \( b = 0.5 \), the exponential equation in the form \( y = ab^x \) is \( y = 3(0.5)^x \).

Answer:

\( y = 3(0.5)^x \)