Question

The axis of symmetry for the function f(x) = –2x2 + 4x + 1 is the line x = 1. Where is the vertex of the function located?

Answer 1

Answer:

(1,3)

Step-by-step explanation:

Answer 2

Answer:

(1, 3)

Step-by-step explanation:

You are given the h coordinate of the vertex as 1, but in order to find the k coordinate, you have to complete the square on the parabola.  The first few steps are as follows.  Set the parabola equal to 0 so you can solve for the vertex.  Separate the x terms from the constant by moving the constant to the other side of the equals sign.  The coefficient HAS to be a +1 (ours is a -2 so we have to factor it out).  Let's start there.  The first 2 steps result in this polynomial:

$-2x^2+4x=-1$.  Now we factor out the -2:

$-2(x^2-2x)=-1$.  Now we complete the square.  This process is to take half the linear term, square it, and add it to both sides.  Our linear term is 2x.  Half of 2 is 1, and 1 squared is 1.  We add 1 into the set of parenthesis.  But we actually added into the parenthesis is +1(-2).  The -2 out front is a multiplier and we cannot ignore it.  Adding in to both sides looks like this:

$-2(x^2-2x+1)=-1-2$.  Simplifying gives us this:

$-2(x^2-2x+1)=-3$

On the left we have created a perfect square binomial which reflects the h coordinate of the vertex.  Stating this binomial and moving the -3 over by addition and setting the polynomial equal to y:

$-2(x-1)^2+3=y$

From this form,

$y=-a(x-h)^2+k$

you can determine the coordinates of the vertex to be (1, 3)