Exponential graphs

What an exponential graph is

An exponential function has the variable in the power, e.g. $y = 2^x$ or $y = a \times b^{\,x}$ .
The graph either grows ever more steeply or decays toward zero, depending on the base:
- $b > 1$ ⇒ growth: the curve rises faster and faster to the right.
- $0 < b < 1$ (or a negative power, $y = 2^{-x}$ ) ⇒ decay: the curve falls toward the axis.

Two exponential curves through the point (0,1): growth y equals two to the x, and decay y equals two to the minus x. Both flatten toward the horizontal asymptote y equals zero.

Key features of an exponential

$y$ -intercept: put $x = 0$ . Since $b^0 = 1$ , the graph $y = a \times b^{\,x}$ crosses the $y$ -axis at $y = a$ (for $y = 2^x$ that is $(0, 1)$ ).
Horizontal asymptote: the curve approaches $y = 0$ but never reaches it — there is no $x$ -intercept for $y = a\,b^{\,x}$ .
Always positive: $b^{\,x}$ is never zero or negative, so the range is $y > 0$ ; the domain is all real $x$ .

Modelling growth and decay

Exponentials model anything that changes by a fixed percentage each step: populations, savings with compound interest, or a car losing value.
Write these as $y = a \times b^{\,x}$ $y = a \times b^{x}$ :
- $a$ — the starting amount (when $x = 0$ ).
- $b$ — the multiplier per step: $b = 1.08$ for $+8\%$ growth, $b = 0.85$ for a $15\%$ loss.

Worked ExampleA decay model

A $24{,}000 car loses $15\%$ of its value each year, so its value after $x$ years is $y = 24000 \times 0.85^{\,x}$ . Find its value when new and after 3 years, and describe the graph.

Step 1 — Value when new ( $x = 0$ )

$y = 24000 \times 0.85^{0} = 24000 \times 1 = \$24{,}000$

This is the $y$ -intercept.

Step 2 — Value after 3 years

$y = 24000 \times 0.85^{3} = 24000 \times 0.614\ldots \approx \$14{,}739$

Step 3 — Describe the graph

Because $b = 0.85 < 1$ , the curve decays: it starts at $24{,}000 and falls, flattening toward the asymptote $y = 0$ — the car keeps losing value but never quite reaches $0.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

State the $y$ -intercept of $y = 5 \times 3^{\,x}$ and whether the graph grows or decays.

Merit

A population of possums is modelled by $P = 200 \times 1.1^{\,t}$ , where $t$ is in years. Find the starting population and the population after 5 years, and state the yearly percentage increase.

Excellence

Two savings options each start at $1000: Option A adds a flat $120 per year ( $y = 1000 + 120x$ ); Option B grows at $10\%$ per year ( $y = 1000 \times 1.1^{\,x}$ ). Determine which is worth more after 10 years, and explain what the graphs show about which option overtakes the other.

Exponential graphs

Covers AS91257

What an exponential graph is

An exponential function has the variable in the power, e.g. $y = 2^x$ or $y = a \times b^{\,x}$ .
The graph either grows ever more steeply or decays toward zero, depending on the base:
- $b > 1$ ⇒ growth: the curve rises faster and faster to the right.
- $0 < b < 1$ (or a negative power, $y = 2^{-x}$ ) ⇒ decay: the curve falls toward the axis.

Two exponential curves through the point (0,1): growth y equals two to the x, and decay y equals two to the minus x. Both flatten toward the horizontal asymptote y equals zero.

Key features of an exponential

$y$ -intercept: put $x = 0$ . Since $b^0 = 1$ , the graph $y = a \times b^{\,x}$ crosses the $y$ -axis at $y = a$ (for $y = 2^x$ that is $(0, 1)$ ).
Horizontal asymptote: the curve approaches $y = 0$ but never reaches it — there is no $x$ -intercept for $y = a\,b^{\,x}$ .
Always positive: $b^{\,x}$ is never zero or negative, so the range is $y > 0$ ; the domain is all real $x$ .

Modelling growth and decay

Exponentials model anything that changes by a fixed percentage each step: populations, savings with compound interest, or a car losing value.
Write these as $y = a \times b^{\,x}$ $y = a \times b^{x}$ :
- $a$ — the starting amount (when $x = 0$ ).
- $b$ — the multiplier per step: $b = 1.08$ for $+8\%$ growth, $b = 0.85$ for a $15\%$ loss.

Worked ExampleA decay model

A $24{,}000 car loses $15\%$ of its value each year, so its value after $x$ years is $y = 24000 \times 0.85^{\,x}$ . Find its value when new and after 3 years, and describe the graph.

Step 1 — Value when new ( $x = 0$ )

$y = 24000 \times 0.85^{0} = 24000 \times 1 = \$24{,}000$

This is the $y$ -intercept.

Step 2 — Value after 3 years

$y = 24000 \times 0.85^{3} = 24000 \times 0.614\ldots \approx \$14{,}739$

Step 3 — Describe the graph

Because $b = 0.85 < 1$ , the curve decays: it starts at $24{,}000 and falls, flattening toward the asymptote $y = 0$ — the car keeps losing value but never quite reaches $0.

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

State the $y$ -intercept of $y = 5 \times 3^{\,x}$ and whether the graph grows or decays.

Merit

A population of possums is modelled by $P = 200 \times 1.1^{\,t}$ , where $t$ is in years. Find the starting population and the population after 5 years, and state the yearly percentage increase.

Excellence

Exponential graphs

What an exponential graph is

Key features of an exponential

Modelling growth and decay

Step 1 — Value when new (x=0x = 0x=0)

Step 2 — Value after 3 years

Step 3 — Describe the graph

Test yourself

Practice by grade

Exponential graphs

What an exponential graph is

Key features of an exponential

Modelling growth and decay

Step 1 — Value when new (x=0x = 0x=0)

Step 2 — Value after 3 years

Step 3 — Describe the graph

Test yourself

Practice by grade

Step 1 — Value when new ( $x = 0$ )

Step 1 — Value when new ( $x = 0$ )