Question

Math help plz! Giving brainlist :D

Answer 1

-x^2+4x-4

dy/dx=-2x+4, d2y/dx2=-2

Since acceleration is a constant negative, when dy/dx, velocity, equals zero, it is at the absolute maximum value for f(x).

dy/dx=0 when -2x+4=0, 2x=4, x=2

f(2)=-x^2+4x-4=-4+8-4=0

So there is a maximum of zero

...

.5x^2+8x+5

dy/dx=x+8, d2y/dx2=1

Since d2y/dx2, acceleration is a constant positive, when dy/dx=0, it is at the point when f(x) is at an absolute minimum.

dy/dx=0 when x+8=0, x=-8

f(-8)=.5x^2+8x+5=32-64+5=-27

So there is a absolute minimum of -27

...

-2(x+1)^2+5

dy/dx=-4(x+1)=-4x-4,  d2y/dx2=-4

dy/dx=9 when -4x-4=0, -4x=4, x=-1

f(-1)=5 

So the vertex is (-1, 5) and the axis of symmetry is the vertical line x=-1