# Week- 8 Finding the vertex using y=a(x−p)2+q

This I have learned how to find the vertex using

y= $a(x-p)^2+q$

Finding the vertex is very important to be able to graph. The vertex tells you the y-intercept, the axis of symmetry, and the range and domain. So first, lets put some numbers instead of P and Q.

y= $a(x-8)^2+7$

When you have this equation, your vertex will always be your P and Q, which is very interesting to me because it can also directly tell you your axis of symmetry. In this case it will x=8. Why didn’t I say x=-8? It is because if you use -8, the equation will have 2 negative which will become positive.  As a result, negative numbers will become positive, and positive number will become negative for this equation.

Let’s add something more to this. If we were told that our line will cross (3,6) and will need to find a, what do we have to do?

It is very simple. Since x=3 and y=6, you plot in the equation y= $a(x-p)^2+q$

6= $a(3-8)^2+7$

6= $a(-5)^2+7$

6= 25a+7

-1=25a

a= $\frac{-1}{25}$

It is that simple. Finding P and Q will find your vertex amazed me. Hopefully there will have more lesson that will amaze me more.