Year 10AC9M10A05

Sketching a Parabola and Transformations

Experiment with functions and relations using digital tools, making and testing conjectures and generalising emerging patterns.

Four features fix any parabola. Once you know how each parameter moves the curve, you can sketch without a table of values.

Method

General form

y = ax^2 + bx + c

$c$ is the $y$-intercept.

Axis of symmetry

x = -\frac{b}{2a}

The turning point sits on it.

Turning point form shows the vertex

y = a(x-h)^2 + k \Rightarrow (h,k)

What each parameter does

a \text{ stretches} \quad h \text{ shifts across} \quad k \text{ shifts up}

Beware: $(x-h)$ shifts right by $h$.

e.g.

\text{Sketch } y = x^2 - 4x + 3

Read the $y$ -intercept from $c$ . $(0, 3)$
Find the axis of symmetry. $x = -\frac{-4}{2} = 2$
Substitute to find the turning point. $y = 4 - 8 + 3 = -1 \ \Rightarrow\ (2,-1)$
Find the $x$ -intercepts by factorising. $(x-1)(x-3) = 0 \ \Rightarrow\ x = 1, 3$
Plot and draw.
Check the roots are symmetric about $x = 2$ . $1 \text{ and } 3 \quad\checkmark$

Intercepts and axis of symmetry first. The turning point follows.

\text{Axis of symmetry of } y = x^2 + 6x + 5

Answer

x = -3

\text{Turning point of } y = (x+2)^2 - 5

Answer

(-2, -5)

\text{Which way does } y = -2x^2 + 1 \text{ open?}

Answer

\text{Downward.}

\text{How does } y = x^2 + 4 \text{ differ from } y = x^2?

Answer

\text{Shifted 4 up.}

\text{How does } y = (x-3)^2 \text{ differ from } y = x^2?

Answer

\text{Shifted 3 right.}

\text{Why does } (x-h) \text{ shift right, not left?}

Answer

\text{The vertex is where the bracket is zero, at } x = h.