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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