SOLUTION: How to find a Parabola (these are some i cannot get right): 1. y=x^2-8x+13 2.y+2x^2+32=-16x-1 and 3. 2x-y^2=4y+10

Question 555513: How to find a Parabola (these are some i cannot get right):
1. y=x^2-8x+13
2.y+2x^2+32=-16x-1
and 3. 2x-y^2=4y+10
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
How to find a Parabola (these are some i cannot get right):
..
1. y=x^2-8x+13
complete the square:
y=(x^2-8x+16)+13-16
y=(x-4)^-3
This is an equation of a parabola which opens upwards and has a vertex at (4,-3)
Standard form of equation: y=A(x-h)^2+k, (h,k) being the (x,y) coordinates of the vertex.
A is a multiplier which affects the slope or steepness of the curve.
..
2.y+2x^2+32=-16x-1
y=-2x^2-16x-33
complete the square:
y=-2(x^2+8x+16)-33+32
y=-2(x+4)^2-1
This is an equation of a parabola which opens downwards and has a vertex at (-4,1)
Standard form of equation: y=-A(x-h)^2+k, (h,k) being the (x,y) coordinates of the vertex.
..
3. 2x-y^2=4y+10
2x=y^2+4y+10
divide by 2
x=(1/2)y^2+2y+5
complete the square:
x=(1/2)(y^2+4y+4)+5-2
x=(1/2)(y+2)^2+3
This is an equation of a parabola which opens rightwards and has a vertex at (3,-2)
Standard form of equation: x=A(y-k)^2+h, (h,k) being the (x,y) coordinates of the vertex.
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