Question 1203974
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The average rate of change (AROC) is the same as the slope of the line through endpoints (-3,h(-3)) and (a,h(a))


If x = -3, then,
h(x) = x^2+3x+2
h(-3) = (-3)^2+3(-3)+2
h(-3) = 2


If x = a, then,
h(x) = x^2+3x+2
h(a) = a^2+3a+2


{{{slope = rise/run}}}


{{{slope = matrix(1,3,"change","in","y")/matrix(1,3,"change","in","x")}}}


{{{slope = (h(a)-h(-3))/(a-(-3))}}}


{{{slope = ((a^2+3a+2)-(2))/(a+3)}}}


{{{slope = (a^2+3a)/(a+3)}}}


{{{slope = (a(a+3))/(a+3)}}}


{{{slope = (a*cross((a+3)))/(cross((a+3)))}}}


{{{slope = a}}}


From here we conclude that <font color=red>a = -1</font> must be the case if we want the slope to be -1.


The average rate of change of h(x) = x^2+3x+2 on the interval {{{-3 <= x <= -1}}} is -1


Graph
{{{graph(400,400,-7,7,-7,7,x^2+3x+2,-x-1)}}}
Parabola h(x) = x^2+3x+2 in red
The line y = -x-1 is in green
The line passes through the points (-3,2) and (-1,0); both of which are on the parabola.
This line has a slope of -1 to represent the AROC on the interval {{{-3 <= x <= -1}}}
Desmos and GeoGebra are two graphing options I recommend.


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Answer: <font color=red size=4>-1</font>
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