y=-f(x+5)+4

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asked Dec 19, 2012 by anonymous
Sketch the parabola y=-f(x+5)+4 where the original parent function isf(x)=x^2 . Label the vertex, roots, y-intercept, and axis of symmetry.

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answered Dec 20, 2012 by anonymous
f=x^2

y=-f(x+5)+4

y=-(x+5)^2+4

parabola that kerv down

y=-x^2 -10x -21...set =0 & get (x+3)(x+7)=0

so y=0 at x=-3 & x=-7

dy/dx =-2x-10, set =0 tu get horizontal tangent...at x=-5

the top av parabola
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