Question

Write an exponential function whose graph passes through the points $\left(0,-2\right)\text{and}\left(-2,-32\right)$.

Short answer

The exponential function passing through the given points is $y=-2{\left(\frac{1}{4}\right)}^x$.

Step-by-step solution

Step 1. Write down the given information.

The given points are $\left(0,-2\right)\text{and}\left(-2,-32\right)$.

Step 2. Concept used.

The exponential function is of the form $y=a{b}^x$ where $a\ne 0,b\ne 1\text{and}b>0$.

Step 3. Satisfying the given points in exponential function.

On satisfying the points $\left(0,-2\right)\text{and}\left(-2,-32\right)$in exponential function $y=a{b}^x$.

$\begin{array}{l}y=a{b}^x\mathrm{....}\left(1\right)\\ \left(-2\right)=a{b}^{\left(0\right)}\mathrm{....}\left(\text{Forthepoint}\left(0,-2\right)\right)\\ a=-\mathrm{2....}\left(2\right)\end{array}$

Similarly,

$\begin{array}{l}y=a{b}^x\\ \left(-32\right)=a{b}^{\left(-2\right)}\mathrm{....}\left(\text{Forthepoint}\left(-2,-32\right)\right)\\ -32=\frac{a}{b^2}\\ -32=\frac{\left(-2\right)}{b^2}\mathrm{....}\left(\begin{array}{l}\text{From}\left(2\right)\\ a=-2\end{array}\right)\\ b=\frac{1}{4}\mathrm{....}\left(\because b>0\right)\end{array}$

Plugging $a=-2,b=\frac{1}{4}$ in (1) gives $y=-2{\left(\frac{1}{4}\right)}^x$ which is the required exponential function passing through the given points.

Step 4. Conclusion.

The exponential function passing through the given points is $y=-2{\left(\frac{1}{4}\right)}^x$.