Question

Write an exponential function whose graph passes through the points $\left(0,3\right)\text{and}\left(-1,6\right)$.

Short answer

The exponential function passing through the given points is $y=3{\left(\frac{1}{2}\right)}^x$.

Step-by-step solution

Step 1. Write down the given information.

The given points are $\left(0,3\right)\text{and}\left(-1,6\right)$.

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 $\left(0,3\right)\text{and}\left(-1,6\right)$in exponential function $y=a{b}^x$.

$\begin{array}{c}y=a{b}^x\mathrm{....}\left(1\right)\\ \left(3\right)=a{b}^{\left(0\right)}\mathrm{....}\left(\text{Forthepoint}\left(0,3\right)\right)\\ 3=a\\ a=\mathrm{3....}\left(2\right)\end{array}$

Similarly,

$\begin{array}{c}y=a{b}^x\\ \left(6\right)=a{b}^{\left(-1\right)}\mathrm{....}\left(\text{Forthepoint}\left(-1,6\right)\right)\\ 6=\frac{a}{b}\\ b=\frac{a}{6}\\ b=\frac{\left(3\right)}{6}\mathrm{....}\left(\begin{array}{l}\text{From}\left(2\right)\\ a=3\end{array}\right)\\ b=\frac{1}{2}\end{array}$

Plugging $a=3,b=\frac{1}{2}$ in (1) gives $y=3{\left(\frac{1}{2}\right)}^x$ which is the required exponential function passing through the given points $\left(0,3\right)\text{and}\left(-1,6\right)$.

Step 4. Conclusion.

The exponential function passing through the given points is $y=3{\left(\frac{1}{2}\right)}^x$