Question

Find the minimum y-value of the vertex.
f(x)=x^2+2x-1
and
f(x)=x^2-5x
Please show work I would like to know how you got the answer to better understand it by someone explaining it.

Answer 1

the min y value occurs at the vertex, so do you know how to find the vertex?

Answer 2

y = a(x – h)2 + k?

Answer 3

that's vertex form

Answer 4

do you know how to find the x coordinate of the vertex?

Answer 5

apparently not, I'm given these forms and examples and have to figure out what i can from them :/

Answer 6

the x coordinate of y = ax^2 + bx + c is

x = -b/(2a)

Answer 7

y = x^2+2x-1 is the same as y = 1x^2+2x-1

in this case, a = 1 and b = 2, so ...

x = -b/(2a)

x = -2/(2*1)

x = -2/2

x = -1

Answer 8

this means that the x coordinate of the vertex for x^2+2x-1 is x = -1

Answer 9

you then plug that back into y = x^2+2x-1 to get the y coordinate of the vertex

Answer 10

another way is to convert it to vertex form

\(y = a(x – h)^2 + k?\)

vertex = (h, k)

Answer 11

f(x)=x^2+2x-1
= (x+1)^2 -1 - 1
= (x--1)^2 -2

compare this with vertex form

vertex = (-1, -2)

Answer 12

I got -25/4 for x^2+2x-1, correct?

Answer 13

no unfortunately it isn't

Answer 14

f(x)=x^2-5x
x=-b/2a
=(-)-5/2*1
=5/2
x=f(5/2)
=(5/2)^2-5*(5/2)
=-25/4?

Answer 15

oh you're working on a different function altogether

Answer 16

you got it correct