How do you graph quadratics using the vertex form?

1 Answer
Dec 26, 2014

When you graph a quadratic, there are a couple of things you need to consider that will make your life easier.

Firstly, you need to find the vertex . This is derived through (wait for it) vertex form. In the equation #y=a(x-h)^2+k#, the point #(h,k)# is your vertex. It is either the highest or lowest point on your graph, and it is the first thing you need to graph.

For example, consider the equation #y = -2(x-6)^2+2#

Here #h = 6# and #k = 2#, therefore the vertex would be #(6,2)#

Secondly, you need to find the direction of the graph. If your #a# value in the above equation is negative , then the equation opens downwards. If it is positive, then it opens upwards. This will help you get a good picture of what your graph will look like.

Considering our sample equation, #a = -2#, therefore it opens down.

Thirdly, you need to find the intercepts of the graph. If you plug in 0 for #y# in the equation above, and solve for #x#, then you will fnd the coordinates of the two intercepts. These will give you additional points to complete your curve.

Now let's find the zeroes of our sample equation:

#0 = -2(x-6)^2+2#

#-2 = -2(x-6)^2#

#1 = (x-6)^2#

#+-1 = x-6#

#x = 6+-1 = {7, 5}#

Hence our intercepts would be the points #(7,0)# and #(5,0)#

Lastly, you just need to draw a curve through all these three points, and you are done! This is what it would look like:

made on desmos online graphing calculator

If you are required to find more coordinates, the easiest way to get them is to just plug the equation into your calculator, and get more points. If you need to do this by hand, then you'd just plug in more #y#'s and solve for #x#.

Hope that helped :)