Quadratic Equations

Hong Kong

Stage 4 - Stage 5

Lesson

How much information do we need to be able to graph a quadratic function? One way is to find the coordinates of the $x$`x`-intercepts, which we can then use to find the coordinates of the vertex. Our first step to the $x$`x`-intercepts will be to factorise the equation.

There are many techniques for factorising quadratics, these are covered in the factorisation section. We can factorise by:

- Using the highest common factor
- Factor special forms such as the difference of two squares or perfect squares
- Grouping in pairs

Recall that if we have two factors, like $a$`a` and $b$`b`, and we multiply them together so that they equal $0$0, then one of those factors ($a$`a` or $b$`b`) *must* be $0$0. A written solution to a question like this would be similar to the following:

If $a\times b=0$`a`×`b`=0 then $a=0$`a`=0 or $b=0$`b`=0.

Now, let's see why this is so useful when graphing quadratics.

Finding the $x$`x`-intercepts

$x$`x`-intercepts occur when the value of $y$`y` is zero

So finding the $x$`x`-intercepts is the same as solving the quadratic. Refer to the lesson on Physical characteristics of solving an equation for more detail. As we saw when Solving by Factorising, once fully factorised, we set each factor to zero and solve to get the values of the zeros. These values correspond to the x-coordinates of the x-intercepts of the graph of the quadratic function.

Find the $x$`x`-intercepts of the function $y=2x\left(x+6\right)$`y`=2`x`(`x`+6).

**Think**: Finding the $x$`x`-intercepts of this function is the same as finding the roots of the equation $2x\left(x+6\right)=0$2`x`(`x`+6)=0.

**Do**: Set each factor to zero, so either

$2x$2x |
$=$= | $0$0 | or | $x+6$x+6 |
$=$= | $0$0 |

$x$x |
$=$= | $0$0 | $x$x |
$=$= | $-6$−6 |

This curve crosses the $x$`x`-axis at $0$0 and $-6$−6.

All parabolas (the shape of a quadratic function) are symmetric functions. The axis of symmetry of a parabola passes through its turning point. Since parabolas have only one turning point, this point is given a special name: the vertex. This is the maximum or minimum value of a quadratic function.

**The vertex always occurs midway between the two roots, and lies on the axis of symmetry.**

In the example above where we had $x$`x`-intercepts of $0$0 and $-6$−6, the $x$`x`-coordinate of the vertex will occur halfway between $0$0 and $-6$−6.

In some cases you can tell by inspection what the midway point is. In this case we can see that it is $-3$−3. If we can't tell by inspection we can just find the average:

$\frac{x_1+x_2}{2}$x1+x22 |
$=$= | $\frac{0+\left(-6\right)}{2}$0+(−6)2 |

$=$= | $\frac{-6}{2}$−62 | |

$=$= | $-3$−3 |

Now remember that this is only the $x$`x`-coordinate of the vertex. We still need to substitute this value into the quadratic to get the $y$`y`-coordinate:

$f(x)$f(x) |
$=$= | $2x\left(x+6\right)$2x(x+6) |

$f(3)$f(3) |
$=$= | $2\times\left(-3\right)\times\left(-3+6\right)$2×(−3)×(−3+6) |

$=$= | $-6\times3$−6×3 | |

$=$= | $-18$−18 |

So the vertex has the coordinates $\left(-3,-18\right)$(−3,−18).

Now that we know that the function $y=2x\left(x+6\right)$`y`=2`x`(`x`+6) has $x$`x`-intercepts at $x=0$`x`=0 and $x=-6$`x`=−6, and has a vertex at $\left(-3,-18\right)$(−3,−18), we can graph the parabola that passes through these three points. The parabola looks like this:

Consider the parabola $y=x\left(x+6\right)$`y`=`x`(`x`+6).

Find the $y$

`y`value of the $y$`y`-intercept.Find the $x$

`x`values of the $x$`x`-intercepts.Write all solutions on the same line separated by a comma.

State the equation of the axis of symmetry.

Find the coordinates of the turning point.

Turning point $=$=$\left(\editable{},\editable{}\right)$(,)

Plot the graph of the parabola.

Loading Graph...

Consider the parabola $y=\left(x+1\right)\left(x-3\right)$`y`=(`x`+1)(`x`−3).

Find the $y$

`y`value of the $y$`y`-intercept.Find the $x$

`x`values of the $x$`x`-intercepts.Write all solutions on the same line separated by a comma.

State the equation of the axis of symmetry.

Find the coordinates of the vertex.

Vertex $=$=$\left(\editable{},\editable{}\right)$(,)

Graph the parabola.

Loading Graph...

Consider the equation $y=x^2-6x+8$`y`=`x`2−6`x`+8.

Factorise the expression $x^2-6x+8$

`x`2−6`x`+8.Hence or otherwise find the $x$

`x`values of the $x$`x`-intercepts of the quadratic function $y=x^2-6x+8$`y`=`x`2−6`x`+8. Write all solutions on the same line separated by a comma.Find the $x$

`x`value of the turning point.Find the $y$

`y`value of the turning point.Plot the graph of the function.

Loading Graph...