Sketching Graphs of the Form <em>y = ax² + q</em>

Sketch the graph of \(y={2x}^{2}-4\). Mark the intercepts and turning point.

Examine the standard form of the equation

We notice that \(a>0\). Therefore the graph is a “smile” and has a minimum turning point.

Calculate the intercepts

For the \(y\)-intercept, let \(x=0\):

\begin{align*} y& = 2{x}^{2}-4 \\ & = 2{(0)}^{2}-4 \\ & =-4 \end{align*}

This gives the point \((0;-4)\).

For the \(x\)-intercepts, let \(y=0\):

\begin{align*} y& = 2{x}^{2}-4 \\ 0& = 2{x}^{2}-4 \\ {x}^{2}& = 2 \\ \therefore x& = ±\sqrt{2} \end{align*}

This gives the points \((-\sqrt{2};0)\) and \((\sqrt{2};0)\).

Determine the turning point

From the standard form of the equation we see that the turning point is \((0;-4)\).

Plot the points and sketch the graph

Domain: \(\{x:x\in \mathbb{R}\}\)

Range: \(\{y:y\ge -4,y\in \mathbb{R}\}\)

The axis of symmetry is the line \(x=0\).

Sketch the graph of \(g(x)=-\cfrac{1}{2}{x}^{2}-3\). Mark the intercepts and the turning point.

Examine the standard form of the equation

We notice that \(a<0\). Therefore the graph is a “frown” and has a maximum turning point.

Calculate the intercepts

For the \(y\)-intercept, let \(x=0\):

\begin{align*} g(x)& = -\cfrac{1}{2}{x}^{2}-3 \\ g(0)& = -\cfrac{1}{2}{(0)}^{2}-3 \\ & = -3 \end{align*}

This gives the point \((0;-3)\).

For the \(x\)-intercepts let \(y=0\):

\begin{align*} 0& = -\cfrac{1}{2}{x}^{2}-3 \\ 3& = -\cfrac{1}{2}{x}^{2} \\ -2(3)& = {x}^{2} \\ -6& = {x}^{2} \end{align*}

There is no real solution, therefore there are no \(x\)-intercepts.

Determine the turning point

From the standard form of the equation we see that the turning point is \((0;-3)\).

Plot the points and sketch the graph

Domain: \(x\in \mathbb{R}\).

Range: \(y\in (-\infty ;-3]\).

The axis of symmetry is the line \(x=0\).