Question

"find the equation of the tangent line at x=a

y = x 3 - 4x - 5; a = 2
A) y=-5
B) y = 8x - 21
C) y = 8x - 5
D) y = 3x - 21

Answer 1

B). y = 8x - 21

Step-by-step explanation:

y = x^3 - 4x - 5

Finding the derivative

y' = 3x^2 - 4

When x = 2

y' = 12 - 4 = 8

- that is the slope of the tangent at x = 2 is 8.

At x = 2, y = 2^3 - 4(2) - 5 = -5.

So we have the equation of the tangent to be

y - y1 = m(x - x1)

--> y - (-5) = 8(x - 2)

--> y + 5 = 8x - 16

--> y = 8x - 16 - 5

--> y = 8x - 21.

Answer 2

To find the equation of the tangent line at x=2 for the curve y=x^3 - 4x - 5, we calculate the derivative to find the slope and then use the point-slope form with the given x value. The equation of the tangent line is y=8x-21.

Step-by-step explanation:

To find the equation of the tangent line to the curve y = x^3 - 4x - 5 at x=a, where a=2, we first need to find the derivative of the curve, which will give us the slope of the tangent line at any point x. Then, we plug in the value of a into the derivative to get the slope of the tangent line at x=2.

Let's find the derivative of y = x^3 - 4x - 5:

y' = 3x^2 - 4
Now, we calculate the slope of the tangent line at x=2:

y'(2) = 3(2)^2 - 4 = 12 - 4 = 8
The slope of the tangent line is 8. To find the equation of the tangent line, we use the point-slope form:

y - y1 = m(x - x1)

Where (x1, y1) is a point on the tangent line, and m is the slope. We know that x1 = 2 and we can find y1 by plugging x1 into the original equation:

y1 = (2)^3 - 4(2) - 5 = 8 - 8 - 5 = -5

Therefore, the equation of the tangent line using the point (2, -5) and slope 8 is:

y + 5 = 8(x - 2)

Simplifying this, we get:

y = 8x - 16 - 5

y = 8x - 21

So the equation of the tangent line is y = 8x - 21, which corresponds to option B.