SOLUTION: Hello, I have to use 'completing the square' to put this equation into standard form, find the vertex, x- and y-intercepts.
{{{2x^2+16x+9}}}
I completed the square like this
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-> SOLUTION: Hello, I have to use 'completing the square' to put this equation into standard form, find the vertex, x- and y-intercepts.
{{{2x^2+16x+9}}}
I completed the square like this
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Question 367153: Hello, I have to use 'completing the square' to put this equation into standard form, find the vertex, x- and y-intercepts.
I completed the square like this:
+/-
From here, I'm not sure how to put in standard form, but I know that once I have it in standard form, the vertex is (h,k).
Didn't I also find the x-intercept when I made the equation equal to zero to complete the square?
I plugged 0 in for x and solved for y to get y-intercept = 9.
If you could please show me how to finish completing the square to put in standard form, and also how I find the x-intercept I would greatly appreciate it! Thanks so much for your valuable time :) Answer by ewatrrr(24785) (Show Source):
Hi,
Not sure exactly what the focus is here:
finding the vertex
The vertex form of a parabola, where(h,k) is the vertex
y =
y = 2[(x^2 + 8x + 9/2)]
y = 2[(x+4)^2 -16 + 9/2)]
y = 2(x+4)^2 + 2*(-23/2)
y = 2(x+4)^2 - 23
vertex (-4,-23)
