Let f(x) =πx4−8x3+ 2x−11. Determine an equation of the tangent line to the graph of f at x=−2. ( line should pass through the same point as f at x=−2, as well as have the same slope as f at x=−2.)

Answer 1

I assume that f(x) = pi*x^4 - 8*x^3 + 2*x - 11.

To find the slope take the derivative:

f'(x) = 4*pi*x^3 - 24*x^2 + 2

At x = -2 the slope is f'(-2) = -32*pi - 94

We can write a straight line in slope-intercept form:

y = m*x + b

where m is the slope and b is the y intercept.

Since the line and f(x) come together at x = -2 we get

f(-2) = f'(-2)*(-2) + b

16*pi + 49 = (-32*pi - 94) * (-2) + b

from which we get

b = -48*pi - 139

So, the line we want is:

y = (-32*pi - 94)*x - 48*pi - 139

Answer 2

Let bfirst get y-coordinate y of tangent point: y = f(-2) = π(-2)⁴ - 8(-2)³ + 2(-2) - 11 = 16π + 49.

Tangent point is (-2, 16π + 49).

To get slope of tangent line let find y' = 4πx³ - 24x² + 2, then slope m = y'(-2) = 4π(-2)³- 24(-2)² - 11 = - 32π - 107.

Then point-slope equation of a line gives: y - (16π + 49) = (-32π - 107)(x - (-2)), or

y = -(32π + 107)x - 64π - 214 + 16π + 49,

at lust y = -(32π + 107)x - 48π - 165