Question

Consider the function f(x) = 6x + x^2 and the point P(-2, -8) on the graph of f.Find the slope of each secant line: (line passing through Q(–3, f(x))) (line passing through Q(–2.5, f(x))) (line passing through Q(–1.5, f(x)))Use the results of part (b) to estimate the slope of the tangent line to the graph of f at P(-2, -8).

Answer 1

Given the function

`f(x)=6x+x^2`

And the point (-2, -8). Therefore:

line passing through Q(–3, f(x))

We have that x = -3, so

`f(-3)=6(-3)+(-3)^2=-18+9=-9`

Then, the slope of line between (-2,-8) and (-3,-9) is:

`\text{slope}=(-9-(-8))/(-3-(-2))=(-9+8)/(-3+2)=(-1)/(-1)=1`

Answer: slope = 1

line passing through Q(–2.5, f(x))

x = -2.5, therefore

`f(-2.5)=6(-2.5)+(-2.5)^2=-15+6.25=-8.75`

The slope of line between (-2,-8) and (-2.5, -8.75) is:

`\text{slope}=(-8.75-(-8))/(-2.5-(-2))=(-8.75+8)/(-2.5+2)=(-0.75)/(-0.5)=1.5`

Answer: slope = 1.5

line passing through Q(–1.5, f(x))

x = -1.5, then

`f(-1.5)=6(-1.5)+(-1.5)^2=-9+2.25=-6.75`

The slope of line between (-2,-8) and (-1.5, -6.75) is:

`\text{slope}=(-6.75-(-8))/(-1.5-(-2))=(-6.75+8)/(-1.5+2)=(1.25)/(0.5)=2.5`

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:

`f^(\prime)(x)=6+2x`

Then substitute x = -2

`f(-2)=6+2(-2)=6-4=2`

Answer: slope = 2