Question

Find the vertex of y=x^2+4x-5. show your work.

Answer 1

1) take the derivative
\[y = x^{2} + 4x - 5\]
=>
\[y' = 2x + 4\]
2) set the derivative equal to zero and solve for x
\[2x+4=0 => x=-2\]
3) plug the x you found back into the original equation
\[y = (-2)^2 + 4(-2) - 5 = -9\]
4) you have both the x and y coordinate now.

Answer 2

how did you get from y=x^2+4x-5 to y=2x+4?

Answer 3

@chrismoon

Answer 4

It's y', not y. Have you learned about derivatives yet?

Answer 5

no.. i'm confused.

Answer 6

Check this page out. Should help:
http://www.purplemath.com/modules/sqrvertx.htm

Answer 7

tthanks.

Answer 8

you could also complete the square. longer route, but good for algebra students :)