Question

j

Answer 1

PLEASE HELP ME!!!

Answer 2

The x-intercepts are where the graph cuts the x-axis. To get on the x-axis, the y-coordinate has to be 0.
So, take y = -x^2 + 12x - 28
y = -x^2 + 12x - 28 and set y = 0.
0 = -x^2 + 12x - 28
x^2 - 12x + 28 = 0
Will the expression factor?

Answer 3

Is this quadratic formula or completing the square? And idk! I'm horrible at math! I wish I understood it better, but I just don't..

Answer 4

>Is this quadratic formula or completing the square? Neither. Those are techniques to find the roots (x-intercepts). The stuff above is getting the equation ready to be solved.

Answer 5

Okay... so now that the equation is ready to be solved.. how do I go about solving it?

Answer 6

The first thing I do is to see if the expression will factor: x^2 - 12x + 28 = 0
Will it?

Answer 7

No?

Answer 8

Okay. Let's use the Quadratic Formula to crank out the roots.

Answer 9

Okay.

Answer 10

Why not the completing the square one, why is the quadratic formula the better way to go in this case?

Answer 11

Completing the square is fine. It is how the Quadratic Formula was created. Do you want to complete the square?

Answer 12

No, we can just do what you originally said, the quadratic formula. I was just wondering. Carry on, sorry.

Answer 13

y = -x^2 + 12x - 28
0 = -x^2 + 12x - 28

a = -1
b = 12
c = -28

Answer 14

What comes next?

Answer 15

I'm not sure..

Answer 16

x^2-12+36-8=0
(x-8)^2-8=0
x=2sqrt2+8 or
x=-2sqrt2+8

Answer 17

typo..

Answer 18

(x-6)^2-8=0
x=2sqrt2+6 or
x=-2sqrt2+6