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.

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

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

a) Well, it seems like you've already answered this one pretty well! The equation of the secant line joining the points (-2, -6) and (4, 0) is indeed y = x - 4. Way to go!

b) Good job taking the derivative! Now, let's set it equal to the slope of the secant line. The slope of the secant line is (0 - (-6))/(4 - (-2)) = 1. Plug this value into the derivative equation: 2x - 1 = 1. Solve for x to find the point c.

c) Now that you have the x-coordinate of the point c, substitute it back into the original function f(x) = x^2 - x - 12 to find the corresponding y-coordinate. Then use the point-slope form of a line to find the equation of the tangent line. You've got this!

a) To find the equation of the secant line joining the points (-2,-6) and (4,0), we can use the slope-intercept form of a linear equation, which is y = mx + b, where m is the slope and b is the y-intercept.

The slope of the secant line can be calculated using the formula:

m = (y2 - y1) / (x2 - x1),

where (x1, y1) and (x2, y2) are the coordinates of the two given points.

For the points (-2,-6) and (4,0), the slope of the secant line is:

m = (0 - (-6)) / (4 - (-2)) = 6 / 6 = 1.

Now, let's find the y-intercept, b. We can use one of the given points - let's use (-2,-6) - and substitute its coordinates into the slope-intercept form:

-6 = 1 * (-2) + b,
-6 = -2 + b,
b = -6 + 2,
b = -4.

Therefore, the equation of the secant line joining the points (-2,-6) and (4,0) is:

y = x - 4.

b) To use the Mean Value Theorem (MVT) to determine a point c in the interval (-2,4) such that the tangent line at c is parallel to the secant line, we need to check if there exists a value c such that the derivative of the function at c is equal to the slope of the secant line, which is 1.

First, let's find the derivative of the original function f(x) = x^2 - x - 12. Taking the derivative of f(x) gives us:

f'(x) = 2x - 1.

Next, we need to solve the equation f'(c) = 1 for c.

2c - 1 = 1,
2c = 2,
c = 1.

Therefore, there exists a point c = 1 in the interval (-2,4) such that the tangent line at c is parallel to the secant line.

c) To find the equation of the tangent line through c, we need to find the slope of the tangent line at c using the derivative of the function at c (which we found to be c = 1).

The slope of the tangent line at c can be calculated using the derivative:

m = f'(c) = 2(1) - 1 = 1.

Now, we have the slope of the tangent line at c, and we also know the point (1, f(1)). We can substitute the values into the point-slope form of a linear equation, which is:

y - y1 = m(x - x1),

where (x1, y1) is the given point and m is the slope.

Using the point (1, f(1)) = (1, -12), we can write the equation of the tangent line at c as:

y - (-12) = 1(x - 1),
y + 12 = x - 1,
y = x - 13.

Therefore, the equation of the tangent line through c = 1 is:

y = x - 13.