Question

The quadratic function f(x) = x^2-Kx+2K has a vertex whose y-coordinate is 5. Find all the possible values of K.

Answer 1

f(x) = (x^2 - Kx + _____) + 2k - _____

We're just Completing the Square. Fill in the two blanks with the SAME correct expression.

Answer 2

Completing the square is where you take the middle term and divide it by 2.

Answer 3

b^2-4ac
(k)^2-4(1)(2k-5)
k^2-8k+40 solve for k

Answer 4

so you did it with the quadratic formula?

Answer 5

Why are we doing all that? Just complete the square!

Answer 6

-1 +- sqrt(1^2-4(1)(2)/2(1)

Answer 7

I don't know how to complete the square...

Answer 8

I have used the quadratic formula almost every time

Answer 9

f(x) = (x^2 - Kx + (K/2)^2) + 2K - (K/2)^2

The y-coordinate of the Vertex is 2K - (K/2)^2

Solve: 2K - (K/2)^2 = 5

Answer 10

K - (K/2)^2 = 5/2

Answer 11

K - K = 5/2

Answer 12

So I divide 2 by all the terms and then square root both sides

Answer 13

It's a quadratic equation. You've solved dozens of these.

2K - (K/2)^2 = 5

2K - (K^2)/4 = 5

8K - K^2 = 20

K^2 - 8K + 20 = 0

Abmon98 had the right idea, but also a little error there, right at the end.

Answer 14

So now I need to solve for the K values?

Answer 15

It is in quadratic form

Answer 16

It doesn't factor and has no real solutions