Find the Exponential Function (2,-12) and (3,-24)

$(2, - 12)$ and $(3, - 24)$

Substitute the values from the given points into the formula for an exponential function, $y = a b^{x}$ .

$- 12 = a b^{2}$

$- 24 = a b^{3}$

Solve the two equations for $a$ and $b$ .

Solve for $a$ in $- 12 = a b^{2}$ .

Rewrite the equation as $a b^{2} = - 12$ .

$a b^{2} = - 12$

$- 24 = a b^{3}$

Divide each term in $a b^{2} = - 12$ by $b^{2}$ and simplify.

Divide each term in $a b^{2} = - 12$ by $b^{2}$ .

$\frac{a b^{2}}{b^{2}} = \frac{- 12}{b^{2}}$

$- 24 = a b^{3}$

Simplify the left side.

Cancel the common factor of $b^{2}$ .

Cancel the common factor.

$\frac{a b^{2}}{b^{2}} = \frac{- 12}{b^{2}}$

$- 24 = a b^{3}$

Divide $a$ by $1$ .

$a = \frac{- 12}{b^{2}}$

$- 24 = a b^{3}$

$a = \frac{- 12}{b^{2}}$

$- 24 = a b^{3}$

$a = \frac{- 12}{b^{2}}$

$- 24 = a b^{3}$

Simplify the right side.

Move the negative in front of the fraction.

$a = - \frac{12}{b^{2}}$

$- 24 = a b^{3}$

$a = - \frac{12}{b^{2}}$

$- 24 = a b^{3}$

$a = - \frac{12}{b^{2}}$

$- 24 = a b^{3}$

$a = - \frac{12}{b^{2}}$

$- 24 = a b^{3}$

Replace all occurrences of $a$ with $- \frac{12}{b^{2}}$ in each equation.

Replace all occurrences of $a$ in $- 24 = a b^{3}$ with $- \frac{12}{b^{2}}$ .

$- 24 = (- \frac{12}{b^{2}}) \cdot b^{3}$

$a = - \frac{12}{b^{2}}$

Simplify the right side.

Cancel the common factor of $b^{2}$ .

Move the leading negative in $- \frac{12}{b^{2}}$ into the numerator.

$- 24 = \frac{- 12}{b^{2}} \cdot b^{3}$

$a = - \frac{12}{b^{2}}$

Factor $b^{2}$ out of $b^{3}$ .

$- 24 = \frac{- 12}{b^{2}} \cdot (b^{2} b)$

$a = - \frac{12}{b^{2}}$

Cancel the common factor.

$- 24 = \frac{- 12}{b^{2}} \cdot (b^{2} b)$

$a = - \frac{12}{b^{2}}$

Rewrite the expression.

$- 24 = - 12 \cdot b$

$a = - \frac{12}{b^{2}}$

$- 24 = - 12 b$

$a = - \frac{12}{b^{2}}$

$- 24 = - 12 b$

$a = - \frac{12}{b^{2}}$

$- 24 = - 12 b$

$a = - \frac{12}{b^{2}}$

Solve for $b$ in $- 24 = - 12 b$ .

Rewrite the equation as $- 12 b = - 24$ .

$- 12 b = - 24$

$a = - \frac{12}{b^{2}}$

Divide each term in $- 12 b = - 24$ by $- 12$ and simplify.

Divide each term in $- 12 b = - 24$ by $- 12$ .

$\frac{- 12 b}{- 12} = \frac{- 24}{- 12}$

$a = - \frac{12}{b^{2}}$

Simplify the left side.

Cancel the common factor of $- 12$ .

Cancel the common factor.

$\frac{- 12 b}{- 12} = \frac{- 24}{- 12}$

$a = - \frac{12}{b^{2}}$

Divide $b$ by $1$ .

$b = \frac{- 24}{- 12}$

$a = - \frac{12}{b^{2}}$

$b = \frac{- 24}{- 12}$

$a = - \frac{12}{b^{2}}$

$b = \frac{- 24}{- 12}$

$a = - \frac{12}{b^{2}}$

Simplify the right side.

Divide $- 24$ by $- 12$ .

$b = 2$

$a = - \frac{12}{b^{2}}$

$b = 2$

$a = - \frac{12}{b^{2}}$

$b = 2$

$a = - \frac{12}{b^{2}}$

$b = 2$

$a = - \frac{12}{b^{2}}$

Replace all occurrences of $b$ with $2$ in each equation.

Replace all occurrences of $b$ in $a = - \frac{12}{b^{2}}$ with $2$ .

$a = - \frac{12}{{(2)}^{2}}$

$b = 2$

Simplify the right side.

Simplify $- \frac{12}{{(2)}^{2}}$ .

Raise $2$ to the power of $2$ .

$a = - \frac{12}{4}$

$b = 2$

Divide $12$ by $4$ .

$a = - 1 \cdot 3$

$b = 2$

Multiply $- 1$ by $3$ .

$a = - 3$

$b = 2$

$a = - 3$

$b = 2$

$a = - 3$

$b = 2$

$a = - 3$

$b = 2$

List all of the solutions.

$a = - 3, b = 2$

Substitute the values found for $a$ and $b$ into $y = a b^{x}$ to find the exponential, ignoring negative numbers for $b$ .

$y = - 3 \cdot 2^{x}$

Find the Exponential Function (2,-12) and (3,-24)

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