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Year 10AC9M10A03
Summary of important concepts

Exponentials, Hyperbolas and Circles

Recognise the connection between algebraic and graphical representations of exponential relations and solve related exponential equations.

Three more curve families. Each has one giveaway feature that identifies it instantly.

Method

Exponential
y=axy = a^x
Passes through $(0,1)$. Has a horizontal asymptote at $y = 0$.
Hyperbola
y=kxy = \frac{k}{x}
Two branches. Asymptotes on both axes. Never touches either.
Circle centred at the origin
x2+y2=r2x^2 + y^2 = r^2
Not a function — it fails the vertical line test.
An asymptote is approached, never reached
This is what makes these curves different from a parabola.
Exponential growth beats any polynomial eventually

Worked example

e.g. Identify and describe y=2x, y=6x and x2+y2=25.\text{Identify and describe } y = 2^x,\ y = \frac{6}{x} \text{ and } x^2 + y^2 = 25.
  1. The first has xx in the index, so it is exponential.y=2xy = 2^x
  2. Find its key point and asymptote.(0,1), asymptote y=0(0,1), \text{ asymptote } y = 0
  3. The second has xx in the denominator, so it is a hyperbola.y=6xy = \frac{6}{x}
  4. State its asymptotes.x=0 and y=0x = 0 \text{ and } y = 0
  5. The third has both variables squared and summed — a circle.x2+y2=25x^2 + y^2 = 25
  6. Read the radius.r=5, centre (0,0)r = 5, \text{ centre } (0,0)

Practice

Look at where xx sits. That identifies the family.

1
What is the y-intercept of y=3x?\text{What is the } y\text{-intercept of } y = 3^x?
Answer11
2
Asymptotes of y=4x?\text{Asymptotes of } y = \frac{4}{x}?
Answerx=0 and y=0x = 0 \text{ and } y = 0
3
Radius of x2+y2=36?\text{Radius of } x^2 + y^2 = 36?
Answer66
4
Solve 2x=32\text{Solve } 2^x = 32
Answerx=5x = 5
5
Why is a circle not a function?\text{Why is a circle not a function?}
AnswerOne x value gives two y values.\text{One } x \text{ value gives two } y \text{ values.}
6
Does y=2x ever reach zero?\text{Does } y = 2^x \text{ ever reach zero?}
AnswerNo. It approaches it forever.\text{No. It approaches it forever.}
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