SOLUTION: what is the vertex form of y=-1/2x^2+4x-14

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Question 683603: what is the vertex form of y=-1/2x^2+4x-14
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
y = -1%2F2x² + 4x - 14

Clear of fractions:

2y = -x² + 8x - 28

Factor -1 out of first two terms on the right:

2y = -1(x² - 8x) - 28

Change the parentheses to brackets so it can contain parentheses:

2y = -1[x² - 8x] - 28

To complete the square inside the bracket:

1.  Take one-half of the coefficient of x.   1%2F2·(-8) = -4
2.  Square the result.                       (-4)² = 16
3.  Add it then subtract it in the bracket:   Add + 16 - 16

2y = -1[x² - 8x + 16 - 16] - 28

Factor the first three terms inside the bracket as a perfect square:

2y = -1[(x-4)² - 16] - 28

Remove the bracket by distributing the -1 leaving the parentheses intact:

2y = -1(x-4)² + 16 - 28

Combine the terms +16 - 28 as -12

2y = -1(x-4)² - 12

Solve for y by dividing through by 2

 y = -1%2F2(x-4)² - 6

Compare that to the standard vertex form:

 y = a(x-h)² + k

We see that the vertex (h,k) is (4,-6) and from the original
equation y = -1%2F2x² + 4x - 14 tells us that the y-intercept
is (0,-14) 



The axis of symmetry is the vertical line through the vertex, and
is therefore the vertical line whose equation is x=-4:



The point that matches the y-intercept on the other side of the
axis of symmetry is (8,-14), so we plot that point:

and we sketch in the parabola:

Edwin