Question

If the tangent line to y=f(x) at (9,−2) passes through the point (5,9), then compute f(9).

Answer 1

The tangent line y = f(x) at (9, - 2) passes through the point (5, 9).

Means that, The tangent line y = f(x) passes through the points (9, - 2) and (5, 9).

 

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

Let the points are (x₁, y₁) = (9, - 2) and (x₂, y₂) = (5, 9).

Slope (m) = [(y₂ - y₁)/(x₂ -x₁)]

m = [(9 - (-2))/(5 - 9)]

m  = [(9 + 2)/(- 4)]

⇒ m = - 11/4.

Thisis the slope of the tangent line.

Now, the tangent line equation is y = (- 11/4)x + b.

 

Find the y - intercept by substituting any point in the tangent line equation say (x, y) = (9, - 2).

- 2 = (- 11/4)(9) + b

b = - 2 + (99/4)

b = (- 8 + 99)/4

⇒ b = 91/4.

The tangent line equation  is y = (- 11/4)x + (91/4).

 

y = f (9) = (- 11/4)(9) + (91/4) = - 99/4 + 91/4 = - 8/4 = - 2.

∴ f (9) = - 2.

Answer 2

Slope = 9+2/5-9 = 11/-4

y = f(x)

Perpendicular line slope = 4/11

line equation is y-9 = 4/11(x-5)

11y-99 =4x-20

11y = 4x-79

y = 4x/11-79/11

f(9) = 4*9/11-79/11

= 36/11-79/11

= -43/11

f(9) = -43/11

y = f(x) at (9,-2)

f(9) = -2