Question

y=x^2-2x-15
how do you find the vertex?

Answer 1

and the x intercepts are at 5 and -3?

Answer 2

begin by completing the square

Answer 3

to find the 'x' coordinate of the vertex,
use x = -b/2a or similarly take the first derivative, set equal to zero and solve for 'x'.
To find the 'y' coord. just plug the 'x' value into the original equation.

Answer 4

Yes, the 'x' intercepts are 5 and -3:
(x+3)(x-5) = 0

Answer 5

and vertex at 1,-16

Answer 6

if you complete the square you'll arrive at \[y=(x-1)^{2}+13\] This is the standard form for a parabola. From this form, we then know that the vertex is at (1, 13)

Answer 7

oops i messed up haha

Answer 8

oops, me too. Sorry

Answer 9

-1,-12?

Answer 10

y=(x−1)2+14
so the vertex is at (1, 14)

Answer 11

obviously the derivative thing isnt working for me ahha

Answer 12

that should work too. Take the first derivative: dy/dx=2x-2
set the first derivative equal to zero: 2x-2=0, x = 1

Answer 13

plug in x= 1 to the original equation to find the y-coordinate of the vertex

Answer 14

I got the coordinates of the vertex to be (1,-16)... After plugging in x=1. I may be wrong.

Answer 15

but when i plug 1 in i get -16?

Answer 16

which is what i had the first time
before i got confused

Answer 17

y(1)=\[y(1) = 1^{2}-2+15=14\]

Answer 18

-15

Answer 19

thats what you were doing wrong jjanelle
haha

Answer 20

1-2+15=14

Answer 21

Oh, janelle, you got the -15 mixed up

Answer 22

hahah thank yall

Answer 23

I can't read, good call.

Answer 24

:)