f(x)=2x^2-12x+9- find vertex,axis of symmtry,two points

Question

given f(x)=2x^2-12x+9  need to find the vertex, axis of symmetry and two points-please help.  I figured the vertex at (3,-9)(  axis of symmetry would be 3?) and that's all I have.

Answer 1

f(x) = 2x^2 - 12x + 9

f(x) = 2(x^2 - 6x + 36) + 9 + 2*36 to complete the square we add 36, but the term is multiplied by 2

f(x) = 2(x - 6)^2 + 81

Therefore:

vertex is (6,81)

axis of symetry is x = 6

y-intercept is (0,81)

to get x-intercept set y = 0

0 = 2(x - 6)^2 +81

2(x - 6)^2 = -81

(x - 6)^2 = -81/2

x - 6 = -81/2         or       x - 6 =  81/2

x = -69/2              or        x = 93/2

(-69/2, 0)              or        (93/2, 0) 

Comments

thank you - my answers are way off.

Answer 2

f(x)=2x^2-12x=27

Answer 3

f(x) = 2x2-12x + 9

Answer 4

Two pionts are (6,9) and (2,-7). You get 9by pluging in 6 for x. Then you get -7 by pluging in 2 for x.

Answer 5

f(x)=2x^2-12x+9

To find the perfect square you can use either b/2 plug your answer in and solve or you can use the formula -b/2a.

-(-12)/2(2) = 3 then you plug this into your original equation 2(3)^2-12(3)+9 = -9

so the vertex is (3,-9)