Answer :
Given the function
[tex]f(x)=6x+x^2[/tex]And the point (-2, -8). Therefore:
line passing through Q(–3, f(x))
We have that x = -3, so
[tex]f(-3)=6(-3)+(-3)^2=-18+9=-9[/tex]Then, the slope of line between (-2,-8) and (-3,-9) is:
[tex]\text{slope}=\frac{-9-(-8)}{-3-(-2)}=\frac{-9+8}{-3+2}=\frac{-1}{-1}=1[/tex]Answer: slope = 1
line passing through Q(–2.5, f(x))
x = -2.5, therefore
[tex]f(-2.5)=6(-2.5)+(-2.5)^2=-15+6.25=-8.75[/tex]The slope of line between (-2,-8) and (-2.5, -8.75) is:
[tex]\text{slope}=\frac{-8.75-(-8)}{-2.5-(-2)}=\frac{-8.75+8}{-2.5+2}=\frac{-0.75}{-0.5}=1.5[/tex]Answer: slope = 1.5
line passing through Q(–1.5, f(x))
x = -1.5, then
[tex]f(-1.5)=6(-1.5)+(-1.5)^2=-9+2.25=-6.75[/tex]The slope of line between (-2,-8) and (-1.5, -6.75) is:
[tex]\text{slope}=\frac{-6.75-(-8)}{-1.5-(-2)}=\frac{-6.75+8}{-1.5+2}=\frac{1.25}{0.5}=2.5[/tex]Answer: slope = 2.5
The slope of the tangent line to the graph of f at P(-2, -8)
The slope of the tangent is equal to the 1st derivative:
[tex]f^{\prime}(x)=6+2x[/tex]Then substitute x = -2
[tex]f(-2)=6+2(-2)=6-4=2[/tex]Answer: slope = 2
