Finding Exponential Equations from Graphs or Data

 

Under certain circumstances, we are able to deduce a function's equation from a few known points the associated curve passes through. For example it may be possible when the form of the equation is known.

Example 1

Suppose we consider the curve of a function with the exponential form $y=a\left(b^x\right)$y=a(bx) which passes through $\left(2,18\right)$(2,18) and $\left(5,486\right)$(5,486). 

These points must satisfy the equation (that is, when substituted, will make the equation true). Thus we know that:

$18$18	$=$=	$ab^2$ab2     (1)
$486$486	$=$=	$ab^5$ab5     (2)

Dividing equation (2) by equation (1)  immediately shows that $b^3=27$b3=27 and thus $b=3$b=3. Thus since $ab^2=18$ab2=18, we have $9a=18$9a=18 and $a=2$a=2. So we identify the function as given by $y=2\left(3\right)^x$y=2(3)x.

 

The information we needed was two points and the form of the equation. More complex forms of the exponential function would require more information.

In the next example, we'll see that, even though there is only one solution, the algebra seems to produce two solutions, until of course the solutions are checked. 

Example 2

Suppose we need to fill in a table for values of a function of the form $y=b^{-x}+c$y=b−x+c as shown here:

$x$x	$-2$−2	$-1$−1	$0$0	$1$1	$2$2
$y$y	$()$()	$5$5	$()$()	$3.5$3.5	$()$()

We know the two points $\left(-1,5\right)$(−1,5) and $\left(1,3.5\right)$(1,3.5), so we form two equations as before:

$5$5	$=$=	$b^1+c$b1+c       $(1)$(1)
$3.5$3.5	$=$=	$b^{-1}+c$b−1+c     $(2)$(2)

This means that from equation (2) we have $c=3.5-\frac{1}{b}$c=3.5−1b​ and so substituting this into equation (1) we have that $5=b+\left(3.5-\frac{1}{b}\right)$5=b+(3.5−1b​) or when simplified $1.5=b-\frac{1}{b}$1.5=b−1b​.

Solving for $b$b we have:

$1.5$1.5	$=$=	$b-\frac{1}{b}$b−1b​
$1.5b$1.5b	$=$=	$b^2-1$b2−1
$3b$3b	$=$=	$2b^2-2$2b2−2
$2b^2-3b-2$2b2−3b−2	$=$=	$0$0
$\left(2b+1\right)\left(b-2\right)$(2b+1)(b−2)	$=$=	$0$0
$b$b	$=$=	$-\frac{1}{2},2$−12​,2

 So there seem to be two possibilities. By substituting back into equation (1) above we find that if $b=-\frac{1}{2}$b=−12​, we find $c=5\frac{1}{2}$c=512​, and for $b=2$b=2 we find $c=3$c=3.

However, the first pairing, delivering $y=\left(-\frac{1}{2}\right)^{-x}+5\frac{1}{2}$y=(−12​)−x+512​ needs to be discarded, because the base b is defined to be a positive number.

So the correct solution becomes $y=2^{-x}+3$y=2−x+3.

Using this rule, we have that, for $x=-2$x=−2, $y=7$y=7, for $x=0$x=0, $y=4$y=4 and for $x=2$x=2, $y=3.25$y=3.25.

 

Further comments

For an simple exponential form like $y=b^x$y=bx, one point will suffice. For example, given the point $\left(2,25\right)$(2,25), we can simply substitute the values into the equation. Thus $25=b^2$25=b2, and $b=5$b=5 (note that $b=-5$b=−5 has to be rejected).

 

 

Worked Examples

Question 1

Question 2

Question 3