I need help with this problem please! If you can, please explain step by step. THank you!

Question

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).

(b) Use the Mean Value Theorem to find a value c in the interval (−2, 4) such that the tangent line to the the function f(x), passing through the point (c, f(c)), is parallel to the secant line from (a).

(c) Find the equation of the tangent line from (b).

Answer 1

a) The points (−2, −6) and (4, 0) are points that are both on the function graph of f(x) = x ² − x − 12 and because of that, they form a secant line to the graph, The slope of that secant line is found using the m = (y₂ - y₁)/(x₂ - x₁) slope equation. Plugging in the points we get:

m = (y₂ - y₁)/(x₂ - x₁)

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

m = 6/6 = 1

b) The Mean Value Theorem says that as long as f(x) is continuous and differentiable on the interval between x = -2 and x = 4, there will be some point (some value of "x") such that the slope of the tangent line is equal to the slope of the secant line in a) above. The slope of the tangent line at any random point is the derivative of f(x). The derivative of f(x) is: f '(x) = 2x - 1 (using the power rule). We ar looking for the value of "x" where the slope is the same as the secant line from a) above so, 2x - 1 = 1, or 2x = 2, or x = 1 so, c = 1,