02.41expgraphs

# Exponential Graphs

Exponential graphs are formed by graphing an exponential function.

An exponential function has the variable in the exponent (or index or power).

## Basic Exponential Graphs

The graph of y = ax (where a is positive but excluding 1) is the basic exponential graph.

The graph of $y = 2^x$ is shown here

The graph of $y = 3^x$ is shown here

The graph of $y = 4^x$ is shown here

All of these exponential graphs have:

• Asympote: $y = 0$
• y-intercept: $(x = 0, \; y = 1)$
• No x-intercepts
• No turning points
• 2nd point at $(1, \; a)$
• Domain: $x \in R$
• Range: $y \in R^+$

These are all strictly increasing graphs (gradient is always positive).

## Exponential Graphs : a > 1

All exponential graphs, y = ax, with a > 1 have the following in common:

• Asymptote at $y = 0$: as $x \rightarrow -\infty, \; y \rightarrow 0^+$
• y-intercept at $(0, \; 1) \; \text{ because } a^0 = 1$
• 2nd point at $(1, \; a) \; \text{ because } a^1 = a$
• Strictly Increasing graphs
• As $x \rightarrow +\infty, \; y \rightarrow +\infty$
• Domain: $x \in R$
• Range: $y \in R^+$

## Exponential Graphs : 0 < a < 1

An exponential graph with fractional a (between 0 and 1) has the same shape but is reflected across the y-axis.

All exponential graphs, y = ax, with 0 < a < 1 have the following in common:

• Asymptote at $y = 0$: as $x \rightarrow +\infty, \; y \rightarrow 0^+$
• y-intercept at $(0, \; 1) \; \text{ because } a^0 = 1$
• 2nd point at $(1, \; a) \; \text{ because } a^1 = a$
• Strictly Decreasing graphs
• As $x \rightarrow -\infty, \; y \rightarrow +\infty$
• Domain: $x \in R$
• Range: $y \in R^+$

## Transformations

We can apply the standard transformations such as dilations, translations and reflections to any exponential graph, y = ax.

#### Example 1

… … Sketch $y = 3 \times 2^{x + 1} - 2$

Method 1

Horizontal asymptote when $2^a = 0$
… … $y = 3 \times 0 - 2$
… … $y = -2$

y-intercept at $x = 0$
… … $y = 3 \times 2^1 - 2$
… … $y = 4 \qquad (0, \; 4)$

{At this point we'd usually look at 21 but we've already done that}

A 2nd point at $2^0 = 1$
… … $x + 1 = 0$
… … $x = -1$

… … $y = 3 × 1 - 2$
… … $y = 1 \qquad (-1, \; 1)$

x-intercept when $y = 0$
… … $3 \times 2^{x + 1} - 2 = 0$

… … $3 \times 2^{x + 1} = 2$

… … … $2^{x + 1} = \frac{2}{3}$

… … {Convert to log form}

… … $x + 1 = \log_2 \big( \frac{2}{3} \big)$

… … $x = \log_2 \big( \frac{2}{3} \big) - 1$

… … {This is an acceptable answer, but we could simplify further}

… … $x = \log_2 \big( 2 \big) - \log_2 \big( 3 \big) - 1$

… … $x = 1 - \log_2 \big( 3 \big) - 1$

… … $x = - \log_2 \big( 3 \big) \qquad \big( -\log_2(3), \; 0 \big)$

Method 2

… … $y = 3 \times 2^{x + 1} - 2$

This is the graph of y = 2x with transformations:

• Dilation by a factor of 3 from the x-axis (in the y-direction) {3 × y}
• Translation by 1 unit to the left {x – 1}
• Translation by 2 units down {y – 2}

In matrix form this is

… … $\left[ \begin{matrix} x' \\ y' \\ \end{matrix} \right] = \left[ \begin{matrix} 1&0 \\ 0&3 \\ \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ \end{matrix} \right] + \left[ \begin{matrix} -1 \\ -2 \\ \end{matrix} \right]$

or

… … $x' = x - 1$
… … $y' = 3y - 2$

{just checking we have that correct:

… … $y = 2^x$

… … $y' = 3y - 2$

… …$y' = 3\big( 2^x \big) - 2$

… … $x = x' + 1$

… … $y' = 3\big( 2^{x' + 1} \big) - 2$

So use
… … $x' = x - 1$
… … $y' = 3y - 2$

Asymptote: $y = 0 \longrightarrow y' = -2$

Original y-intercept: $(x = 0, \; y = 1) \longrightarrow (x' = -1, \; y' = 1)$

Original 2nd point: $(x = 1, \; y = 2) \longrightarrow (x' = 0, \; y' = 4)$

Locate x-intercept as shown above.

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