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Answer:
- 10x +y +17 = 0
- (-1/2, -5/8), (1/2, 5/8)
Step-by-step 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)
