Finding the Vertex of a Parabola: A Quick Guide

[jeetp26]

Homework Statement

Find the vertex of the parabola y = (a-b)(a+b)

Homework Equations

x = -b/2a

The Attempt at a Solution

This question was extra credit on my Pre-Calc test today. I got the answer and it took almost a page to do it. But I'm very anxious and I just can't wait until i get my test back. Anyways, I thought if anyone out there would try it and see if i get the same answer.

 

[danago]

Homework Statement

Find the vertex of the parabola y = (a-b)(a+b)

Homework Equations

x = -b/2a

The Attempt at a Solution

This question was extra credit on my Pre-Calc test today. I got the answer and it took almost a page to do it. But I'm very anxious and I just can't wait until i get my test back. Anyways, I thought if anyone out there would try it and see if i get the same answer.

Are you sure that you've written the question exactly as it was stated in the exam? There is no x in the equation you have posted, yet your answer makes reference to an x-coordinate?

 

[HallsofIvy]

I thought at first that you meant y= (x- a)(x- b) but that has vertex at x= (a+b)/2, not -b/2a.

However, you should be able to recognize -b/2a as the part of the quadratic formula outside the square root: The roots of $ax^2+ bx+ c= 0$ are
$$\frac{-b\pm\sqrt{b^2- 4ac}}{2a}$$.

That equation will have exactly one root- that is, the vertex will lie on the x-axis if $b^2- 4ac}= 0$ and, in that case, the root (and so vertex) is given by x= -b/(2a). Since changing c just "moves" the graph up and down, the x-coordinate of the vertex will always be at x= -b/(2a).

The vertex of $y= ax^2+ bx+ c$ is at x= -b/(2a).

 

[QuarkCharmer]

We area always told to express things like this (the vertex) as a coordinate pair.


$$\frac{-b}{2a} = X_{v}$$
(ie: Only the x coordinate of the vertex)