Quadratic Equations

Lesson

Let's review what we know about graphing quadratic functions. Quadratic functions can be written in each of the following forms:

- $y=ax^2+bx+c$
`y`=`a``x`2+`b``x`+`c`- This is known as the general form or $y$`y`-intercept form. - $y=a\left(x-d\right)\left(x-e\right)$
`y`=`a`(`x`−`d`)(`x`−`e`) - This is known as the factored or $x$`x`-intercept form. - $y=a\left(x-h\right)^2+k$
`y`=`a`(`x`−`h`)2+`k`- This is known as the turning point form.

Each of these names gives us a clue to the various features we can identify on a parabola.

When graphing a quadratic, no matter in which form it is presented, we must find the $x$`x`-intercepts, the $y$`y`-intercept, and the turning point.

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 parabola $y=x^2-6x+8$`y`=`x`2−6`x`+8.

a. Determine the value of the $y$`y`-intercept.

The $y$`y`-intercept is located along the $y$`y`-axis, where $x=0$`x`=0.

Substituting $x=0$`x`=0 into our function gives us $y=0^2-6\times0+8$`y`=02−6×0+8.

The $y$`y`-intercept is $8$8.

b. Determine the coordinates of the $x$`x`-intercepts.

It is easiest to find the $x$`x`-intercepts by first factorising our quadratic function. If we cannot factorise, we might like to use the quadratic formula instead.

Factorising gives us $y=\left(x-4\right)\left(x-2\right)$`y`=(`x`−4)(`x`−2).

And as we saw in the first example, this gives us $x$`x`-intercepts of $(4,0)$(4,0) and $(2,0)$(2,0).

c. Determine the coordinates of the turning point.

We begin by finding the $x$`x` value of the turning point and there are a number of ways to do this. Either we can use the vertex formula or we can simply find the midpoint between the two $x$`x`-intercepts.

Halfway between $(4,0)$(4,0) and $(2,0)$(2,0) is $x=3$`x`=3.

To the find the $y$`y` value, we substitute $x=3$`x`=3 back into our equation (and we're welcome to use the form given to us or the factorised version we created in part (b)).

$y=3^2-6\times3+8$`y`=32−6×3+8

$y=-1$`y`=−1

Now that we have all three features, we can plot the points on a graph and join them with a smooth curve.

Consider the parabola $y=-\left(x+1\right)^2+4$`y`=−(`x`+1)2+4.

a. Determine the value of the $y$`y`-intercept.

Once again, we find the $y$`y`-intercept by substituting $x=0$`x`=0 into our function.

$y=-\left(0+1\right)^2+4$`y`=−(0+1)2+4

$y=3$`y`=3

b. Determine the coordinates of the $x$`x`-intercepts.

The easiest way to do this is to solve $0=-\left(x+1\right)^2+4$0=−(`x`+1)2+4.

$0$0 | $=$= | $-\left(x+1\right)^2+4$−(x+1)2+4 |

$-4$−4 | $=$= | $-\left(x+1\right)^2$−(x+1)2 |

$4$4 | $=$= | $\left(x+1\right)^2$(x+1)2 |

$x+1$x+1 |
$=$= | $\pm2$±2 |

$x$x |
$=$= | $1$1 |

$x$x |
$=$= |
$-3$−3 |

So the $x$`x`-intercepts are $1,0$1,0 and $-3,0$−3,0.

c. Determine the coordinates of the turning point.

Since the function is already in turning point form, we simply read it from the equation and we get $-1,4$−1,4.

Now that we have all three features, we can plot the points on a graph and join them with a smooth curve.

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...