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Vectors II

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y = |f(x)| and y = f(|x|)

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Exponentials and logarithms

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y = e^x

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y = mx + c

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Summary

Exponential functions

In a nutshell

Exponential functions are of the form $y=a^x$ where $a$ is a constant greater than $0$ and $x$ is the exponent. Exponential functions are always greater than zero, that is $y=a^x > 0$ for all values of $x$ . Whenever $a>1$ , $y=a^x$ is an increasing function that grows without limit as $x$ increases. Whereas, for $0<a<1$ , $y=a^x$ is a decreasing function that grows without limit as $x$ decreases.

Sketching exponentials

Consider the function $y=2^x$ . The table below contains the values of this function evaluated at some integers close to zero.

$x$	$-2$	$-1$	$0$	$1$	$2$
$y$	$\dfrac14$	$\dfrac12$	$1$	$2$	$4$

The graph of $y=2^x$ is a smooth curve that looks like this:

Maths; Exponentials and logarithms; KS5 Year 12; Exponential functions

As $x$ decreases, the value of $2^x$ tends towards zero. The dotted line $y=0$ above is an asymptote for $y=2^x$ . As the value of $x$ increases the function grows without a limit.

Example 1

Sketch the graphs of $y=3^x$  and $y=1.5^x$  on the same axes.

Here the constants $3$  and $1.5$  are greater than $1$ , so both functions are increasing functions that tend to infinity as $x$  increases, and in this case both tend toward $0$  as  $x$  decreases.

For both graphs, $y=1$  when $x=0$ .

When $x>0$ , $3^x > 1.5^x$ .

When $x<0$ , $3^x < 1.5^x$ .

Example 2

Sketch the graphs of $y=\left(\dfrac12\right)^x$ and $y=2^x$ .

Since $0<\dfrac12<1$ , $y=\left(\dfrac12\right)^x$  is a decreasing function that tends to infinty as $x$ decreases, and in this case it tends towards $0$  as $x$  increases.

$y=\left(\dfrac12\right)^x$  is a reflection in the $y$ -axis of the graph of $y=2^x$ .

If $f\left(x\right)=2^x$ , then:

$y=\left(\dfrac12\right)^x = \left(2^{-1}\right)^x = \left(2\right)^{-x} =f\left(-x\right)$

Note: the graph of $y=f\left(-x\right)$  is a relection of the graph of $y=f\left(x\right)$  in the $y$ -axis.

Example 3

Sketch the graph $y=\left(\dfrac13\right)^x+2$ $.$  Find the coordinates of the point where the graph crosses the y-axis and give the function of the asymptote.

If $f\left(x\right)=\left(\dfrac13\right)^x$ , then

$y=\left(\dfrac13\right)^x+2 = f\left(x\right)+2$

The graph is a translation of the graph $y=\left(\dfrac13\right)^x$ upwards by $2$ .

The graph crosses the $y$ -axis when $x=0$ .

Substituting $x=0$  into $y=\left(\dfrac13\right)^x+2$ , gives

$\begin{aligned}y&=\left(\dfrac13\right)^0+2\\&=1+2\\&=3\end{aligned}$

Therefore, the graph crosses the $y$ -axis at $\underline{(0,3)}$ and the asymptote is given by $\underline{y =2}$ .

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