Question

30 -35 points Workings please.

1. Find the equation of the tangent to the curve with equation
3x^2 + 2x - 5 at the point where x = - 2

2. At what point(s) on the curve with equation
y = 5x^3 is the tangent parallel to the line with equation 15x - 4y + 8 = 0 ?

Thank you for any help.
30 points (there are other questions differentiation similar on my profile.)

Answer 1

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Answer:

1. 10x +y +17 = 0
2. (-1/2, -5/8), (1/2, 5/8)

Explanation:

1. The slope at any point on the curve can be found by differentiating the function.

f(x) = 3x^2 +2x -5

f'(x) = 6x +2 . . . . . derivative of f(x)

f'(-2) = 6(-2) +2 = -10

The value of the function at x=-2 is ...

f(-2) = 3(-2)^2 +2(-2) -5 = 3

So, we want the equation of the line with slope -10 through point (-2, 3). In point-slope form, that equation is ...

y -3 = -10(x +2) . . . . . . . point-slope equation of the tangent

10x +y +17 = 0 . . . . . . . tangent line in general form

__

2. The slope of the given line can be found by solving for y.

4y = 15x +8 . . . . . . add 4y

y = 15/4x +2 . . . . . divide by 4 . . . . . slope is 15/4

The curve has slope ...

y' = 15x^2

These slopes are equal when ...

15/4 = 15x^2

1/4 = x^2

±1/2 = x

The corresponding y-values are ...

y = 5(±1/2)^3 = ±5/8

The points where the tangent to the curve is parallel to the given line are ...

(-1/2, -5/8), (1/2, 5/8)