y= -x2-8x-16

Question

I need to know whether the graph opens up or down, the axis of symmetry, the vertex, the y-intercept, the x-intercepts

Answer 1

y=-x^2 -8x -16

-x^2 tell yu its a parabola that open DOWN, axis vertikal

tu get zeors  (x-intersepts) set y=0 or x^2 +8x +16=0

zeroes...x=-4 & x=-4

tu get the "vertex" get slope=differensheate...

dy/dx=-2x -8=0, or x=8/-2=-4

this tell yu the top av kerv is at x=-4 & y=0....(-4,0)

axis av prabola gotta go thru this point,  so axis=x=-4

y-intersept: set x=0...y=-16

Answer 2

125x^3+175x^2-200x-280