SOLUTION: Find the vertex and than graph 2x^2+8x+16=0 By using: X= -b/2a

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Question 1038677: Find the vertex and than graph
2x^2+8x+16=0
By using:
X= -b/2a


Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
You should study what is presented here, very closely:
Change a quadratic equation from general form to standard form - video

You can directly read the vertex coordinates from the standard form equation. That gives information for graphing.

Observe that your given equation is equivalent to x%5E2%2B4x%2B8=0, so you should use this as your starting general-form equation.
--
xCoordinateofVertex=-b%2F%282a%29=-4%2F%282%2A1%29=-2------Actually this is for the function y=x%5E2%2B4x%2B8. The vertex occurs at x=-2.

The only realistic graph for x%5E2%2B4x%2B8=0 would be TWO points on a single numberline, only if these are real numbers and not containing imaginary parts. The zeros for your equation are .

These are NOT real numbers.

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

Find the vertex and than graph
2x^2+8x+16=0
By using:
X= -b/2a
2x%5E2+%2B+8x+%2B+16+=+0
2%28x%5E2+%2B+4x+%2B+8%29+=+2%280%29 -------- Factoring out GCF, 2
x%5E2+%2B+4x+%2B+8+=+0


Vertex: matrix%281%2C3%2C+%22%28-+2%22%2C+%22%2C%22%2C+%224%29%22%29
graph%28300%2C300%2C-15%2C15%2C-+14%2C100%2Cx%5E2+%2B+4x+%2B+8%29