# Quadratic graphs

## In a nutshell

Quadratic graphs have equations of the form $y=ax^2+bx+c$ and, when plotted, are curved like a $\cup$ or like an $\cap$. Solving a quadratic equation using the corresponding quadratic graph is a case of identifying the $x$-intercepts.

##

## Prerequisites

This summary assumes you are comfortable with at least one of the following methods: factorising quadratics; completing the square of a quadratic; using the quadratic formula.

## What do quadratic graphs look like?

Quadratic graphs have either a $\cup$-shape (a positive quadratic, on the left below) or a $\cap$-shape (a negative quadratic, on the right below):

## What do their equations look like?

Quadratic equations have the form

$y=ax^2+bx+c$

where $a$, $b$ and $c$ are constant numbers and $a$ is not $0$.

If $a$ is positive, then the graph will be $\cup$-shaped. If $a$ is negative, then it will be $\cap$-shaped. This is a quick, first indication as to what your quadratic curve will look like based on its equation.

**Note:** Unlike with linear equations, there is not a constant value for the gradient. This is because the gradient is not constant *on a quadratic curve*.

For example:

$y=x^2+5x+6$

This is the equation of a quadratic graph, where $a=1$, $b=5$ and $c=6$.

## Identifying the $x$-intercepts

When looking for $x$-intercepts on any graph, set $y$ in the equation to $0$ and work out the corresponding $x$-value(s). Hence, to find the $x$-intercepts of the graph of $y=ax^2+bx+c$, you solve:

$0=ax^2+bx+c$

This can be achieved using one of the three methods outlined in the prerequisites at the top.

##### Example 1

*Find the $x$-intercepts of the curve with equation *

*$y=x^2+5x+6$. *

*First set $y=0$, then factorise: *

*$0=x^2+5x+6\\0=(x+3)(x+2)$*

*Hence*

*$x=-3$ and $x=-2$*

*These give the two $x$-intercepts:*

*$\underline{(-3,0)}$ and $\underline{(-2,0)}$*

## Drawing a quadratic graph

Without using graphing software, only a rough sketch of a quadratic curve is enough. There are important features of the graph that you can get right and use in your sketch:

- the shape: either $\cup$-shaped or $\cap$-shaped. This is based on the $a$-term in $y=ax^2+bx+c$;
- the $y$-intercept: this is the $c$-term in $y=ax^2+bx+c$;
- the $x$-intercept(s): sometimes there are two, sometimes one and sometimes none. Solving $0=ax^2+bx+c$ find these.

Once you have established these three characteristics of the quadratic curve, you can give a fairly accurate sketch of it.

#### PROCEDURE

**1.**
| Ensure the equation is in the form $y=ax^2+bx+c$. Rearrange to this if necessary. |

**2.** | Identify the shape of the quadratic curve. If $a>0$ then it is $\cup$-shaped, if $a<0$, then it is $\cap$-shaped. |

**3.** | Identify the $y$-intercept. This is the $c$-term in the equation of the curve. Mark this point on: $(0,c)$. Your curve will pass through it. |

**4.** | By solving $ax^2+bx+c=0$ (either by factorising, completing the square or using the quadratic formula), identify the $x$-intercept(s). Mark however many there are (zero, one or two) on the $x$-axis. If there are no solutions and hence no $x$-intercepts, then you may have to use graph transformations to draw the curve. |

**5.**
| Join the points in the correct curve. |

##### Example 2

*Draw the curve that has equation*

*$y=x^2+5x+6$*

*For this equation, $a=1$ which is positive, so this curve is $\cup$-shaped. Since $c=6$, the *$y$*-intercept is at *$(0,6)$*. Finally, as given in the example above, the *$x$*-intercepts are at *$(-2,0)$* and *$(-3,0)$*. Combining these gives the following sketch:*

## Solving quadratic equations using a graph

To find the $x$-intercept(s) of a quadratic curve, you have to solve a quadratic equation. This works the other way too: if you have the graph of $y=ax^2+bx+c$, then the $x$-intercepts give the solutions to the quadratic equation $ax^2+bx+c=0$.