SOLUTION: Write the vertex form for a parabola with the given characteristics. 1. vertex ( 0, 0) directrix x = -15 2. vertex (3, 3) focus (3, 0) 3. Vertex ( 0, 0 ) focus ( 2, 0)

Question 664345: Write the vertex form for a parabola with the given characteristics.
1. vertex ( 0, 0) directrix x = -15
2. vertex (3, 3) focus (3, 0)

3. Vertex ( 0, 0 ) focus ( 2, 0)
Write the standard form for the parabola given the following equation.
4. x2 + 8x - y + 20 = 0

Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Write the vertex form for a parabola with the given characteristics.
1. vertex ( 0, 0) directrix x = -15
This is a parabola that opens rightwards
Its form of equation: (y-k)^2=4p(x-h)
p=15 (distance from vertex to directrix)
4p=60
equation:y^2=60x
..
2. vertex (3, 3) focus (3, 0)
This is a parabola that opens downwards
Its form of equation: (x-h)^2=-4p(y-k)
p=3 (distance from vertex to focus)
4p=12
equation:(x-3)^2=-12(y-3)
3. Vertex ( 0, 0 ) focus ( 2, 0)
This is a parabola that opens rightwards
Its form of equation: (y-k)^2=4p(x-h)
p=2 (distance from vertex to focus)
4p=8
equation:y^2=8x
..
Write the standard form for the parabola given the following equation.
4. x2+8x-y+20=0
standard form of equation for a parabola: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex, A=multiplier that affects the slope or width of the curve. For A>0 parabola opens upwards, for A<0, parabola opens downwards.
y=x^2+8x+20
complete the square:
y=(x^2+8x+16)+20-16
y=(x+4)^2+4
This is an equation of a parabola that opens upwards with vertex at (-4,4)
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