Question

What is the vertex of y=2x^2-8x+6?

Answer 1

the easiest way in which we can find the vertex is to rewrite the given y=2x^2-8x+6 in the form y-k =a(x-h)^2, where (h,k) is the vertex.

y=2x^2-8x+6 can be rewritten as y = 2(x^2 - 4x ) + 6

Completing the square, y = 2(x^2 - 4x +4 - 4 ) + 6

Then y = 2(x-2)^2 - 2. Comparing this to a(x-h)^2, we see that h=2 and k=2. Thus the vertex is at (2, 2).

Answer 2

The vertex is [ 2, 6]

Explanation:

To find the vertex, all we need to do is to find;

[-b/2a , f(-b/2a)]

From the equation;

y=2x²-8x+6

comparing the above equation by the standard equation

y = ax² + bx + c

a = 2 b = -8 and c = 6

We can now proceed to insert the values

x = -b/2a = 8/2(2) = 8/4 = 2

Next is to find f(-b/2a)

To find f(-b/2a), we are going to replace x by our value of -b/2a =2

f(x) = 2x²-8x+6

f (-b/2a) = 2(2)² -8(2) + 6

=2(4) - 8(2) + 6

=8 - 8 + 6

=6

Therefore the vertex is [ 2, 6]