Please help ! Math
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f(x) = 2(x - 2)^2 - 1

vertex: (2, -1)

Step-by-step explanation:

f(x) = 2x^2 - 8x + 7

First, we find the vertex.

x = -(-8)/4 = 2

y = 2(2)^2 - 8(2) + 7 = 2(4) - 16 + 7 = 8 - 9 = -1

vertex: (2, -1)

Second, we write f(x) in vertex form.

we know that h and k have to be 2 and -1.

2x^2 - 8x + 7 = a(x - 2)^2 - 1

Since 8 - 1 = 7, we do this:

f(x) = (2x^2 - 8x + 8) - 1

Factor out the 2 and then factor the polynomial

f(x) = 2(x^2 - 4x + 4) - 1

factors of 4:

1 4

2 2

-1 -4

-2 -2 = -4

the function in vertex for is:

f(x) = 2(x - 2)^2 - 1

the answer is a if my teacher taught me correct

y = 3x+4

step-by-step explanation:

to find the slope

slope = (y2-y1)/ (x2-x1)

    = (-8-4)/ (-4-0)

    = -12/-4

  = 3

using the slope intercept form of a line

y= mx+b

we know the slope is 3

we know the y intercept is 4 ( the point where it crosses the y axis or x=0)

y = 3x+4