consider the graph of the function f(x)= x^2-x-12

Question

a) find the equation of the secant line joining the points (-2,-6) and (4,0)

 

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

c) find an equation of the tangent line through c

Answer 1

 The curve f(x) = x² - x - 12

a) To find secant line equation.

Let the joining points are (x₁, y₁) = (- 2, -6) and (x₂, y₂) = (4, 0).

Slope-intercept form line equation is y = mx + b, where m is slope and b is y-intercept.

Slope (m) = [(y₂ - y₁)/(x₂ -x₁)]

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

m  = [6/6]

⇒ m = 1.

Now, the line equation is y = (1)x + b.

Find the y - intercept by substituting any point in the line equation say (x, y) = (- 2,-6).

- 6 = (1)(- 2) + b

b = - 6 + 2

b = - 4

⇒ b = -4

Secant line is y = x - 4.

b) The point (-2 , 4)

f(b) = (4)² - 4 - 12 = 16 - 4 - 12 = 0

f(a) = (-2)² + 2 - 12 = 4 + 2 - 12 = -6

From mean value theorem

f(x) = x² - x - 12

f '(x) = 2x  - 1

f '(x) = m

2x - 1 = 1

2x = 1 + 1

2x = 2

x = 1

Substitute x value in the curve f(x) = x² - x - 12

f(x) = (1)² - 1 - 12

= 1 - 1 - 12

= -12

Tangent point c = (x, y) = (1, -12).

c) To find the tangent line equation, substitute the values of m = 1 and (x, y ) = ( 1, -12)  in the slope intercept form of an equation y = mx + b.

-12 = 1 + b

b = - 12  - 1

b = - 13

Substitute m = 1 and b = -13 in y = mx + b.

Tangent line equation is y = x - 13.