Question 1183710

The instantaneous rate of change at {{{x = 2}}} is zero.
Means: A peak trough or inflection. The curve levels off there; the tangent is parallel to the x-axis.

The instantaneous rate of change at {{{x = 3}}} is negative.
Means: There's a downward slop at {{{x=3}}}

The average rate of change on the interval{{{ 0 <=x <=4}}} is zero.
Means: It's symmetric over the interval - for every up there is a down, and you end as high as you start.

Thus: It's a hill. Plot a parabola with a peak at {{{x=2}}}.

or, plot a curve between points:

({{{0}}},{{{0}}}) ,({{{1}}},{{{3}}}) ,({{{2}}},{{{4}}}) ,({{{3}}},{{{3}}}) ({{{4}}},{{{0}}})


<a href="https://ibb.co/grsmLmy"><img src="https://i.ibb.co/grsmLmy/download.png" alt="download" border="0"></a>


you can also find equation using the points above


{{{y=ax^2+bx+c}}}..........({{{0}}},{{{0}}})

{{{0=a*0^2+b*0+c}}}
{{{c=0}}}

{{{3=a*1^2+b*1}}} ..........({{{1}}},{{{3}}})
{{{3=a+b}}}............eq.1 


{{{4=a*2^2+b*2 }}}..............({{{2}}},{{{4}}})
{{{4=4a+2b}}}......divide by {{{2}}}

{{{2=2a+b}}}......eq.2 

from eq.1 and eq.2 we have system

{{{a+b-3=0}}}
{{{2a+b-2=0}}}
------------------------------
{{{2a+2b-6=0}}}...........multiply by {{{2}}}
{{{2a+b-2=0}}}
------------------------subtract
{{{2a+2b-6-2a-b+2=0}}}
{{{b-4=0}}}
{{{b=4}}}


{{{3=a+b}}}............eq.1 
{{{3=a+4}}}
{{{a=3-4}}}
{{{a = -1 }}}

{{{y=-x^2+4x}}}

vertex | ({{{2}}}, {{{4}}})


check the average rate of change of {{{f(x)=-x^2+4x}}} on the interval [{{{0}}},{{{4}}}]

the average rate of change of {{{f(x)}}} on the interval [{{{a}}},{{{b}}}] is 
{{{(f(b)-f(a))/(b-a)}}}

we have that {{{a=0}}}, {{{b=4}}}

{{{f(0)=-0^2+4*0=0}}}
{{{f(4)=-4^2+4*4=0}}}

thus, {{{(f(b)-f(a))/(b-a)=(0-0)/(4-0)=0}}}


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