Question

I need help finding the axis of symmetry.

f(x)= −2(x − 4)2 + 2

Answer 1

@E.ali

Answer 2

is that -2(x-4)^(2) + 2

Answer 3

Yes, sorry about that

f(x)= −2(x − 4)^2 + 2

Answer 4

given a vertex: (p,q) the axis of symmetry is defined as x=p

you have a vertex form of a quadratic .... can you spot the x part?

Answer 5

if you expand it all out to the ax^2 + bx + c form .... the axis can be defined as: -b/(2a)

Answer 6

2x-8^2 + 2

x=2

-b/(2a) = -8/(2)2

-8/(2)2 = -8/4

-8/4=-2

Answer 7

i think your calcs are in error

Answer 8

square first, then distribute

Answer 9

−2(x − 4)^2 + 2

−2(x^2-8x +16) + 2

etc ...

Answer 10

-b/(2a) = -16/(2)2

-16/(2)2 = -16/4

-16/4=-4?

Answer 11

−2(x^2-8x +16) + 2

-2x^2-16x-32+2?

Answer 12

-2x^2-16x-30

-4x-16x-30

-20x-30?

Answer 13

@amistre64

Answer 14

-2x^2-16x-30 this is good, now -b = --16
2a = 2(-2)

what does that simplify to?

Answer 15

-b=16

2a=-4

Answer 16

doh!! -2(-8) = 16 .... not -16

Answer 17

-16/-4 = 4

Answer 18

-2x^2 +16x -30 this is good, now -b = -16
2a = 2(-2)

-16
---- = 4, thats the axis
-4

Answer 19

Thanks, could you help me find the axis of symmetry for g(x) = 5x2 − 10x + 7 as well?