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Question 716641: Please help me solve this equation: 
a.Write the equation in standard form.
b.Find the coordinates of the vertex and focus,direction of opening, the equations for the directrix and the axis of symmetry,and latus rectum
c.Graph the equation of the parabola
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! solve this equation:
a.Write the equation in standard form.
4x2+48x+y+158=0
y=-4x^2-48x-158
..
b.Find the coordinates of the vertex and focus,direction of opening, the equations for the directrix and the axis of symmetry,and latus rectum.
complete the square:
y=-4(x^2+12x+36)-158+144
y=-4(x+6)^2-14 (vertex form of equation, y=-A(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex.)
(y+14)=-4(x+6)^2
(x+6)^2=-(1/4)(y+14) (basic form of equation, (x-h)^2=4p(y-k), (h,k)=(x,y) coordinates of the vertex.)
parabola opens downward
vertex: (-6,-14)
axis of symmetry: x=-6
4p=1/4
p=1/16
focus: (-6,-(14+(1/16)) (p-distance below vertex on the axis of symmetry)
directrix: y=((14+(1/16) (p-distance above vertex on the axis of symmetry)
latus rectum=focal width=4p=1/4
c.Graph the equation of the parabola
see graph below:
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