Question 609154: Y=2x(squared)-8x-1
i wanna know the a vaule b vaule and c vaule i also wanna know if it opens up or down? the axis of symmetry the vertex the domain and the range
Found 2 solutions by ewatrrr, Edwin McCravy:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi
Using the vertex form of a parabola,  where(h,k) is the vertex
y = 2x^2 - 8x - 1 ||Note the use of ^(uppercase 6) to signify the exponent
y = 2(x-2)^2 - 8 - 1 || completing the Square
y = 2(x-2)^2 - 9 || V(2,-9)min, a = 2>0, Parabola opens Upward, Range is y≥ -9, domain x is (−∞,∞)
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website!
I do it the "vertex formula" way
y = 2x² - 8x - 1
Compare your equation to this model:
y = ax² + bx + c
It's easy to see that a=2, b=-8, c = -1
We know that it opens upward because "a" is a positive number.
The vertex formula is:
The vertex or turning point is (h,k) where h =  and
k = ah²+bh+c and the axis of symmetry is the vertical line whose
equation is x = h
So for your vertex, h =  =  = 2
and k = ah²+bh+c = (2)(2)²+(-8)(2)+(-1) = 2(4)-16-1 = 8-16-1 = -9
So the vertex is (h,k) = (2,-9)
The y-intercept is found by letting x=0 in the original equation:
y = 2x² - 8x - 1
y = 2(0)² - 8(0) - 1
y = -1
The y-intercept is the point (0,-1)
So we plot the vertex the axis of symmetry and the y-intercept:
Now you can draw in the parabola so that it is symmetrical with
the green line of symmetry.
All x-values are in the domain, so the domain is all reals ( ,  )
The only y-values are those from the vertex which is -9 upward,
so the range is (-9,).
Edwin
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