Question

Write an exponential function whose graph passes through the points $\left(0,-5\right)\text{and}\left(-3,-135\right)$.

Short answer

The exponential function passing through the given points is $y=-5{\left(\frac{1}{3}\right)}^x$.

Step-by-step solution

Step 1. Write down the given information.

The given points are $\left(0,-5\right)\text{and}\left(-3,-135\right)$.

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 $\left(0,-5\right)\text{and}\left(-3,-135\right)$in exponential function $y=a{b}^x$.

$\begin{array}{c}y=a{b}^x\mathrm{....}\left(1\right)\\ \left(-5\right)=a{b}^{\left(0\right)}\mathrm{....}\left(\text{Forthepoint}\left(0,-5\right)\right)\\ a=-\mathrm{5....}\left(2\right)\end{array}$

Similarly,

$\begin{array}{c}y=a{b}^x\\ \left(-135\right)=a{b}^{\left(-3\right)}\mathrm{....}\left(\text{Forthepoint}\left(-3,-135\right)\right)\\ \left(-135\right)=\frac{\left(-5\right)}{b^3}\mathrm{....}\left(\begin{array}{l}\text{From}\left(2\right)\\ a=-5\end{array}\right)\\ b=\frac{1}{3}\mathrm{....}\left(\because b>0\right)\end{array}$

Plugging $a=-5,b=\frac{1}{3}$ in (1) gives $y=-5{\left(\frac{1}{3}\right)}^x$ which is the required exponential function passing through the given points.

Step 4. Conclusion.

The exponential function passing through the given points is $y=-5{\left(\frac{1}{3}\right)}^x$.