Question

write an exponential function $y = a {b}^{x}$whose graph passes through the given points

Answer 1

The general form of an exponential function is given as;

$y=ab^x$

Here, we are given two pointa along the curve of the graph. These are;

$(2,24)\text{ and }(3,144)$

We shall substitute the values of x and y into the general form and we'll now have;

$\begin{gathered} y=ab^x \\ \text{Where; }(x,y)=(2,24) \\ 24=ab^2 \end{gathered}$

We now divide both sides by a and we'll have;

$\frac{24}{a}=b^2---(1)$

We do the same for the second set of coordinates and this would result in the following;

$\begin{gathered} y=ab^x \\ \text{Where; }(x,y)=(3,144) \\ 144=ab^3 \\ \text{Divide both sides by a and we'll have;} \\ \frac{144}{a}=b^3---(2) \end{gathered}$

At this point, we shall refine equation (1) and make a the subject of the equation;

$\begin{gathered} \frac{24}{a}=b^2---(1) \\ \text{Cross multiply and we'll now have;} \\ \frac{24}{b^2}=a \\ a=\frac{24}{b^2} \end{gathered}$

We can now substitute for the value of a into equation (2);

$\begin{gathered} \frac{144}{a}=b^3 \\ \text{When } \\ a=\frac{24}{b^2} \\ \frac{144}{(\frac{24}{b^2})}=b^3 \end{gathered}$

The left side of the equation can be re-arranged as follows;

$\begin{gathered} \frac{144}{1}\div\frac{24}{b^2}=b^3 \\ \frac{144}{1}\times\frac{b^2}{24}=b^3 \\ 6b^2=b^3 \end{gathered}$

Now we divide both sides by b^2 and we have;

$\begin{gathered} \frac{6b^2}{b^2}=\frac{b^3}{b^2} \\ 6=b \end{gathered}$

We now have the value of b as 6. we can substitute this into equation (1);

$\begin{gathered} \frac{24}{a}=b^2 \\ \frac{24}{a}=6^2 \\ \frac{24}{a}=36 \\ \text{Cross multiply and;} \\ \frac{24}{36}=a \\ \frac{2}{3}=a \end{gathered}$

The values of a and b have now been calculated.

We can now go back and use these values to write up the exponential function using the general form;

$\begin{gathered} y=ab^x \\ \text{Where,} \\ a=\frac{2}{3},b=6 \\ y=\frac{2}{3}(6)^x \end{gathered}$

ANSWER:

$y=\frac{2}{3}(6)^x$