Question 408258
{{{f(x)=-(4/5)x^2 - (1/5)x+1}}}


The vertex is found using the formula for (x,y):

(-b/2a,f(-b/2a))

find {{{x}}}

{{{x=-b/2a}}}

={{{-(-1/5)/2(-4/5)}}}

={{{0.2/(-1.6)}}}

={{{-0.125}}}...so we have (-0.125,y)..now find {{{y}}}

{{{y=f(-0.125)}}}

{{{f(-0.125)= - (4/5)(-0.125)^2 - (1/5)(-0.125)+1}}}

{{{f(1)-0.125=- (4/5)0.015625+0.025+1}}}

{{{f(-0.125)=1.0125}}}................ so  the vertex is  (-0.125,1.0125)



the intervals on which the function is increasing and the intervals on which the function is decreasing

Given the vertex of (-0.125,1.0125):

The function is increasing on:

(-∞,-0.125)

The function is decreasing  on:

(-0.125,∞)



{{{ graph( 500, 500, -10, 10, -10, 10, -(4/5)x^2 - (1/5)x +1) }}}