To graph a quadratic function, it is helpful to know the overall shape (parabola) and a few key points.
The points that are most useful are:
-intercept- vertex
-intercepts (if there are any)- symmetry of the -intercept
These can all be calculated directly from the factored form.
Example
Let’s begin by drawing and axes, and labelling the axis from 
to 
. We leave the axis blank for now until we know how far down or up the vertex is.
The intercept occurs when the coordinate is zero. Therefore when 
:
![\[y=(x-3)(x+5)=(0-3)(0+5)=(-3)(5)=-15\]](https://tentotwelvemath.com/wp-content/ql-cache/quicklatex.com-4ef608f33c64247c62568e227dc182fb_l3.png "Rendered by QuickLaTeX.com")
The -intercept is the point 
.
The -intercepts occur when the coordinate is zero. To locate them, we need to solve the equation
![\[0=(x-3)(x+5)\]](https://tentotwelvemath.com/wp-content/ql-cache/quicklatex.com-6995493dc460da12cad546f63a10d01c_l3.png "Rendered by QuickLaTeX.com")
This turns out to be quite straightforward. When any two numbers multiply to give zero, one or the other or both of the numbers must be zero. Therefore, if 
then either 
or 
.
The -intercepts occur at the points 
and 
.
The coordinate of the vertex is right in the middle of the intercepts. To find the middle of 
and 
, we can do a variety of calculations. One way is to add them together and divide by 2 (find the average of the two numbers).
![\[x_{\text{vertex}}=\dfrac{-5+3}{2}=\dfrac{-2}{2}=-1\]](https://tentotwelvemath.com/wp-content/ql-cache/quicklatex.com-0a8dfc87a40d0aeff9c0524353b1cc55_l3.png "Rendered by QuickLaTeX.com")
Now to find the coordinate of the vertex, we simply substitute the coordinate into the equation for . When 
:
![\[y=(x-3)(x+5)=(-1-3)(-1+5)=(-4)(4)=-16\]](https://tentotwelvemath.com/wp-content/ql-cache/quicklatex.com-8fd11da84f6715293ef30118dd03c780_l3.png "Rendered by QuickLaTeX.com")
![\[y_{\text{vertex}}=-16\]](https://tentotwelvemath.com/wp-content/ql-cache/quicklatex.com-e7e046d2c3473cb5891cb17fd4115bf4_l3.png "Rendered by QuickLaTeX.com")
The coordinates of the vertex are 
.
We now know that the lowest point on the parabola is at 
, and so let’s use a scale of 5 on the axis to accommodate the vertex.
Let’s plot all these points 
on the graph.
The last point to plot is the symmetry of the intercept. If you imagine a mirror line through the vertex, and reflect the intercept over that to plot a fifth point.
We are now ready to draw the smoothest parabola curve through the plotted points, making sure to round off the vertex and not make it too pointy.
Note that a function such as 
or 
will generate a parabola that turns downwards rather than upwards.
Note that if it is not possible to factor a quadratic expression, we must use a different technique for graphing. If there are -intercepts we can find them using the quadratic formula. If there are no -intercepts, then we are left with just the vertex, the -intercept and the symmetry of the -intercept. Since we can’t ‘add the roots then divide by two’ to get the -coordinate of the vertex, we need a different technique. Several techniques to find the vertex are discussed on the page Find the Vertex, which the last page of this unit.
Practice