Question

Find the vertex of the parabola whose equation is y = x^2 + 8x + 12.

Answer 1

the x coordinate of the vertex can be obtained by x = -b/(2a) , where a, b are the coefficients in the standard way of writing the quadratic equation...

Answer 2

x=-b/2a

a=1 b=8 c=12

Answer 3

ax^2 + bx + c = 0...

Answer 4

For any parabola of the form \(ax^2+bx+c=0 \) the vertex will be at

\( \large \left(-\frac b {2a} , -\frac {b^2-4ac}{4a}\right) \).

Answer 5

so -(8)/2*1 = -4

Answer 6

now plug that back in to the quadratic get the value of y if you need it

Answer 7

ok

Answer 8

(-4, 12)
(-4, -4)
(0, -6)

Answer 9

now let me c

Answer 10

have you got it :D

Answer 11

im confused

Answer 12

\[y=x^2+8x+12 = -4\]

wherever you see x put -4 and do the sum

Answer 13

y = (-4)^2 + 8x(-4) + 12 = ??

Answer 14

y = (-4)^2 + 8(-4) + 12 = ??

Answer 15

-36

Answer 16

so
-4x-4 =16
-4x8 = -32
then add 12

16-32+12 =-4

Answer 17

so the vertex is at (-4,-4)

Answer 18

ok thank you