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 factor the equation.
Recall that there are many techniques for factoring quadratics. We can factor by:
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.
$x$x-intercepts occur when the value of $y$y is zero
So finding the $x$x-intercepts is the same as solving the quadratic. As we saw when Solving by Factoring, once fully factored, 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=2x(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$2x(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 |
Reflect: This curve crosses the $x$x-axis at $\left(0,0\right)$(0,0) and $\left(-6,0\right)$(−6,0).
All parabolas (the shape of a quadratic function) are symmetric functions. The axis of symmetry (line 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 where we find the maximum or minimum value of a quadratic function.
The vertex of a parabola occurs midway between the two roots and lies on the axis of symmetry.
In the example above where we had $x$x-intercepts of $x=0$x=0 and $x=-6$x=−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 halfway 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 |
$f(x)$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=2x(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 $y=x\left(x-4\right)$y=x(x−4).
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 $=$=$\left(\editable{},\editable{}\right)$(,)
Draw a graph of the parabola.
Consider the parabola $y=\left(x+2\right)\left(x-4\right)$y=(x+2)(x−4).
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.
Consider the equation $y=x^2+2x-8$y=x2+2x−8.
Factor the expression $x^2+2x-8$x2+2x−8.
Hence or otherwise find the $x$x values of the $x$x-intercepts of the quadratic function $y=x^2+2x-8$y=x2+2x−8. Write all solutions on the same line separated by a comma.
Find the $x$x value of the vertex.
Find the $y$y value of the vertex.
Plot the graph of the function.