Find the Exponential Function (0,-5) and (4,-80)

$(0, - 5)$ and $(4, - 80)$

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

$- 5 = a b^{0}$

$- 80 = a b^{4}$

Simplify the right side.

Simplify $a b^{0}$ .

Anything raised to $0$ is $1$ .

$- 5 = a \cdot 1$

$- 80 = a b^{4}$

Multiply $a$ by $1$ .

$- 5 = a$

$- 80 = a b^{4}$

$- 5 = a$

$- 80 = a b^{4}$

$- 5 = a$

$- 80 = a b^{4}$

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

Rewrite the equation as $a = - 5$ .

$a = - 5$

$- 80 = a b^{4}$

Replace all occurrences of $a$ with $- 5$ in each equation.

Replace all occurrences of $a$ in $- 80 = a b^{4}$ with $- 5$ .

$- 80 = (- 5) \cdot b^{4}$

$a = - 5$

Simplify the right side.

Multiply $- 5$ by $b^{4}$ .

$- 80 = - 5 b^{4}$

$a = - 5$

$- 80 = - 5 b^{4}$

$a = - 5$

$- 80 = - 5 b^{4}$

$a = - 5$

Solve for $b$ in $- 80 = - 5 b^{4}$ .

Rewrite the equation as $- 5 b^{4} = - 80$ .

$- 5 b^{4} = - 80$

$a = - 5$

Divide each term in $- 5 b^{4} = - 80$ by $- 5$ and simplify.

Divide each term in $- 5 b^{4} = - 80$ by $- 5$ .

$\frac{- 5 b^{4}}{- 5} = \frac{- 80}{- 5}$

$a = - 5$

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

$\frac{- 5 b^{4}}{- 5} = \frac{- 80}{- 5}$

$a = - 5$

Divide $b^{4}$ by $1$ .

$b^{4} = \frac{- 80}{- 5}$

$a = - 5$

$b^{4} = \frac{- 80}{- 5}$

$a = - 5$

$b^{4} = \frac{- 80}{- 5}$

$a = - 5$

Simplify the right side.

Divide $- 80$ by $- 5$ .

$b^{4} = 16$

$a = - 5$

$b^{4} = 16$

$a = - 5$

$b^{4} = 16$

$a = - 5$

Take the 4th root of both sides of the equation to eliminate the exponent on the left side.

$b = \pm \sqrt[4]{16}$

$a = - 5$

Simplify $\pm \sqrt[4]{16}$ .

Rewrite $16$ as $2^{4}$ .

$b = \pm \sqrt[4]{2^{4}}$

$a = - 5$

Pull terms out from under the radical, assuming positive real numbers.

$b = \pm 2$

$a = - 5$

$b = \pm 2$

$a = - 5$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$b = 2$

$a = - 5$

Next, use the negative value of the $\pm$ to find the second solution.

$b = - 2$

$a = - 5$

The complete solution is the result of both the positive and negative portions of the solution.

$b = 2, - 2$

$a = - 5$

$b = 2, - 2$

$a = - 5$

$b = 2, - 2$

$a = - 5$

Solve the system of equations $b = 2, a = - 5$ .

$b = 2, a = - 5$

Solve the system of equations $b = - 2, a = - 5$ .

$b = - 2, a = - 5$

List all of the solutions.

$b = 2, a = - 5$

$b = - 2, a = - 5$

$b = 2, a = - 5$

$b = - 2, a = - 5$

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

$y = - 5 \cdot 2^{x}$

Find the Exponential Function (0,-5) and (4,-80)

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