Exponential Functions

New Zealand

Level 7 - NCEA Level 2

Lesson

Any function can be transformed by adding something to it and any function can be transformed by multiplying it by a number. We might write

$f(x)\rightarrow g(x)=f(x)+c$`f`(`x`)→`g`(`x`)=`f`(`x`)+`c`

$f(x)\rightarrow h(x)=af(x)$`f`(`x`)→`h`(`x`)=`a``f`(`x`)

We apply this general principle to exponential functions. Thus, if $f(x)=2^x$`f`(`x`)=2`x` then $g(x)$`g`(`x`) might be $2^x+5$2`x`+5 and $h(x)$`h`(`x`) might be $-\frac{1}{2}\cdot2^x$−12·2`x`.

The function $g(x)=2^x+5$`g`(`x`)=2`x`+5 is just the function $f(x)$`f`(`x`) with $5$5 added to every function value. The graph of $g(x)$`g`(`x`) must look the same as the graph of $f(x)$`f`(`x`) but shifted $5$5 units up the vertical axis. The following diagram shows these two functions.

Observe that $f(0)=1$`f`(0)=1 and $g(0)=1+5=6$`g`(0)=1+5=6, as expected.

The function $f(x)$`f`(`x`) is asymptotic to the horizontal axis and $g(x)$`g`(`x`) is asymptotic to the line $y(x)=5$`y`(`x`)=5.

At every point $x$`x`, the distance between $f(x)$`f`(`x`) and $g(x)$`g`(`x`) is $5$5.

The function $h(x)=-\frac{1}{2}\cdot2^x$`h`(`x`)=−12·2`x`, is the function $f(x)$`f`(`x`) with every function value multiplied by $-\frac{1}{2}$−12.

Multiplication by $\frac{1}{2}$12 brings all values of $f(x)$`f`(`x`) closer to zero by that factor. The graph of $h(x)$`h`(`x`) will appear compressed in the vertical direction compared with the graph of $f(x)$`f`(`x`).

Since all the values of $f(x)$`f`(`x`) are positive, all the values of $h(x)$`h`(`x`) must be negative. That is, the graph of $h(x)$`h`(`x`) is not only compressed in the vertical direction but is also reflected across the horizontal axis.

The graphs are represented in the following diagram.

Observe that $f(0)=1$`f`(0)=1 but $h(0)=-\frac{1}{2}$`h`(0)=−12; $f(1)=2$`f`(1)=2 but $h(1)=-1$`h`(1)=−1; $f(2)=4$`f`(2)=4 but $h(2)=-2$`h`(2)=−2; and so on, as expected.

Answer the following.

Determine the $y$

`y`-intercept of $y=2^x$`y`=2`x`.Hence or otherwise determine the $y$

`y`-intercept of $y=2^x-2$`y`=2`x`−2.Determine the horizontal asymptote of $y=2^x$

`y`=2`x`.Hence or otherwise determine the horizontal asymptote of $y=2^x-2$

`y`=2`x`−2.

Consider a graph of $y=5^x$`y`=5`x`.

How could the graph of $y=-5^x$

`y`=−5`x`be obtained from the graph of $y=5^x$`y`=5`x`?through a vertical translation

Athrough a reflection across the $y$

`y`-axisBthrough a reflection across the $x$

`x`-axisCby making it steeper

DGiven the graph of $y=5^x$

`y`=5`x`, sketch $y=-5^x$`y`=−5`x`on the same coordinate plane.Loading Graph...

This is a graph of $y=3^x$`y`=3`x`.

How do we shift the graph of $y=3^x$

`y`=3`x`to get the graph of $y=3^x-4$`y`=3`x`−4?Move the graph $4$4 units to the right.

AMove the graph downwards by $4$4 units.

BMove the graph $4$4 units to the left.

CMove the graph upwards by $4$4 units.

DHence plot $y=3^x-4$

`y`=3`x`−4 on the same graph as $y=3^x$`y`=3`x`.