Question

what is y=x^2-8x+3 in vertex form???

Answer 1

Complete the square

-2(x^2+4x) - 3

-2(x^2+4x+4-4) - 3

-2(x^2+4x+4) + 8 - 3

-2(x^2+4x+4) + 5

-2(x+2)^2 + 5

Answer 2

Completing the square is one way to do it. Here is another:

y=x^2-8x+3 is in the form y = ax^2 + bx + c where

a = 1
b = -8
c = 3

Use the axis of symmetry formula:

x = -b/(2a)

x = -(-8)/(2*1)

x = 4

This is the x coordinate of the vertex. Plug this into y=x^2-8x+3 to find the y coordinate of the vertex.

y=x^2-8x+3

y=(4)^2-8(4)+3

y = 16 - 8(4) + 3

y = 16 - 32 + 3

y = -13

This is the y coordinate of the vertex.

Therefore, the vertex is (h,k) = (4, -13) which means we know that

a = 1 (given)
h = 4 (just found this)
h = -13 (just found this)

Finally you plug these values into the general vertex equation below to get this

y = a(x-h)^2 + k

y = 1(x-4)^2 + (-13)

y = (x-4)^2 - 13

Answer 3

thank you!

Answer 4

you're welcome