Mathematics » Functions I » Quadratic Functions

# Sketching Graphs of the Form y = ax2 + q

## Sketching Graphs of the Form $$y=a{x}^{2}+q$$

In order to sketch graphs of the form $$f(x)=a{x}^{2}+q$$, we need to determine the following characteristics:

1. sign of $$a$$

2. $$y$$-intercept

3. $$x$$-intercept

4. turning point

## Example

### Question

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 $$x=0$$.

## Example

### Question

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 $$x=0$$.

Do you want to suggest a correction or an addition to this content? Leave Contribution