Question 716641
solve this equation: {{{4x2+48x+y+158=0}}}
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:

{{{ graph( 300, 300, -10, 10, -20, 10,-4x^2-48x-158) }}}