3b in 8 express the function y= 2x2 + 8x + 1 in vertex form. y= a0(x + next question ask for turni

y = 2[x + 2]^2 - 7

Step-by-step explanation:

We want to express y= 2x2 + 8x + 1 in vertex form.

Rewrite y= 2x2 + 8x + 1   as   y= 2(x^2 + 4x)              + 1

Now "complete the square" of x^2 + 4x:

Identify the coefficient of the x term.  it is 4.  

Take half of this:  it is 2.

Square this result (that is, square 2) and then add the result to x^2 + 4x, and then subtract it:  x^2 + 4x becomes x^2 + 4x + 4 - 4.  Convince yourself that x^2 + 4x + 4 - 4 is identical to x^2 + 4x.    

x^2 + 4x + 4 can be rewritten as (x + 2)^2.

Going back to our equation y= 2(x^2 + 4x)              + 1   (see above),

replace "x^2 + 4x" in this equation with "(x + 2)^2 - 4:

Then:    y= 2(x^2 + 4x)              + 1  becomes:

             y= 2( [x + 2]^2 - 4  )          + 1, or

              y = 2[x + 2]^2 - 8  +  1, or    y = 2[x + 2]^2 - 7

Compare this to the standard equation

             y = a(x-h)^2 + k.  We see that h = -2 and k = -7.

The given equation, expressed in vertex form, is y = 2[x + 2]^2 - 7.  The vertex is at (-2, -7).

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