Exponential Functions

New Zealand

Level 7 - NCEA Level 2

Lesson

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.

Suppose we consider the curve of a function with the exponential form $y=a\left(b^x\right)$`y`=`a`(`b``x`) 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 $b^3=27$`b`3=27 and thus $b=3$`b`=3. Thus since $ab^2=18$`a``b`2=18, we have $9a=18$9`a`=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.

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

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.

For an simple exponential form like $y=b^x$`y`=`b``x`, 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=`b`2, and $b=5$`b`=5 (note that $b=-5$`b`=−5 has to be rejected).

Find the equation of the curve in the form $y=a^x$`y`=`a``x`.

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Consider the given table of values.

$x$x |
$1$1 | $2$2 | $3$3 | $4$4 |
---|---|---|---|---|

$y$y |
$10$10 | $100$100 | $1000$1000 | $10000$10000 |

Identify the common ratio between consecutive $y$

`y`values.State the equation relating $x$

`x`and $y$`y`.Find the value of $y$

`y`when $x=10$`x`=10.

Consider the table of values.

number of days passed ($x$x) |
$1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
---|---|---|---|---|---|

population of krill ($y$y) |
$4$4 | $16$16 | $64$64 | $256$256 | $1024$1024 |

Is the number of krill increasing by the same amount each day?

Yes

ANo

BFind the equation linking population, $y$

`y`, and time, $x$`x`, in the form $y=a^x$`y`=`a``x`.

Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs

Apply graphical methods in solving problems