Asked by Ray
                Consider the graph of the function f(x)=x^2-x-12
a) Find the equation of the secant line joining the points (-2,-6) and (4,0).
I got the equation of the secant line to be y=x-4
b) Use the Mean Value Theorem to determine a point c in the interval (-2,4) such that the tangent line at c is parallel to the secant line.
I took the derivative of the original question to be f(x)=2x-1 but don't know what to do from there
c) Find the equation of the tangent line through c.
            
        a) Find the equation of the secant line joining the points (-2,-6) and (4,0).
I got the equation of the secant line to be y=x-4
b) Use the Mean Value Theorem to determine a point c in the interval (-2,4) such that the tangent line at c is parallel to the secant line.
I took the derivative of the original question to be f(x)=2x-1 but don't know what to do from there
c) Find the equation of the tangent line through c.
Answers
                    Answered by
            Ray
            
    b) Using the mean value theorem gave me f'(c)=1 and solving for c gave me 1 as well.
c) Tangent line y=x-13
I checked with a graphing calculator and my answers look right
    
c) Tangent line y=x-13
I checked with a graphing calculator and my answers look right
                    Answered by
            Reiny
            
    your secant equation is correct
checking the other parts:
f '(x) = 2x - 1 ---> you had that
setting that equal to 1, since the slope of our secant is 1
2x-1 = 1
2x = 2
x = 1
so all you have to do is sub that back into the original
f(1) = 1 - 1 - 12 = -12
So the point where the slope of the secant is equal to the slope of the tangent is (1, -12)
the equation of the tangent at that point is
y+12 = 1(x-1)
y = x - 13 ----> you had that
good job.
You seem to be able to take care of the mechanics of the problem, but perhaps understanding what all that means seems a bit fuzzy.
I suggest you take a look at KhanAcademy video of this, especially the first two parts.
(These videos done by Sal Khan are extremely well done)
https://www.khanacademy.org/math/ap-calculus-ab/derivative-applications-ab/mean-value-theorem-ab/v/mean-value-theorem-1
The second part fits your problem extremely close
    
checking the other parts:
f '(x) = 2x - 1 ---> you had that
setting that equal to 1, since the slope of our secant is 1
2x-1 = 1
2x = 2
x = 1
so all you have to do is sub that back into the original
f(1) = 1 - 1 - 12 = -12
So the point where the slope of the secant is equal to the slope of the tangent is (1, -12)
the equation of the tangent at that point is
y+12 = 1(x-1)
y = x - 13 ----> you had that
good job.
You seem to be able to take care of the mechanics of the problem, but perhaps understanding what all that means seems a bit fuzzy.
I suggest you take a look at KhanAcademy video of this, especially the first two parts.
(These videos done by Sal Khan are extremely well done)
https://www.khanacademy.org/math/ap-calculus-ab/derivative-applications-ab/mean-value-theorem-ab/v/mean-value-theorem-1
The second part fits your problem extremely close
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