Question

general question:

Answer 1

f(x)=x^2-4x .... ok so complete the sq etc.. (x-2)^2 -4

Answer 2

so the vertex is 2, -4 , if it had been (x+2)^2 -4 would the vertex have been (-2, -4) - not saying that the equation would work out like that.. just using as an example

Answer 3

in other words.. if it is not in the form (x-h)+k... what do u do?

Answer 4

You want to know the vertex of a quadratic that isn't in vertex form?

Answer 5

i guess?

Answer 6

-b/2a yeah?

Answer 7

The x coordinate is:
\[-b/2a\]

Answer 8

and a function of that is the y?

Answer 9

you can plug that into the equation to get the y coordinate

Answer 10

so if it is (x+2)^2 +4 ... do you need to convert it or can u mathematically say that the vertex is (-2, 4)?

Answer 11

Well, that is in vertex form, so you know that the vertex is at (h,k). Do you want to know why that is?

Answer 12

um.. i guess im just confused when it isnt (x-h) + k... and its (x+h)+k ... do u just make h a negative?

Answer 13

yes, because:\[(x+h)^2+k = (x-(-h))^2+k\]

Answer 14

awesome. just as i was thinking... wish i still had my algebra book.. this precalc book doesnt list that.. and i couldnt remember , thanks so much

Answer 15

You are welcome

Answer 16

If you are wondering why that is:
\[a(x-h)^2 +k = a(x^2 - 2hx + h^2)+k = ax^2 - 2ahx +ah^2 + k\]

Answer 17

we know that the x coordinate is:
\[-\frac{b}{2a} = \frac{2ah}{2a} = h\]

Answer 18

if we plug that into the equation we get:
\[ah^2 + ah^2 - 2ah^2 + k = k\]

Answer 19

which is our y coordinate