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zzz [600]
3 years ago
5

Given f(x)=-2x^3 + 3x^2 , find the equation of that tangent line of f at the point where x=2.

Mathematics
1 answer:
trapecia [35]3 years ago
5 0

Step 1: Find f'(x):

f'(x) = -6x^2 + 6x

Step 2: Evaluate f'(2) to find the slope of the tangent line at x=2:

f'(2) = -6(2)^2 + 6(2) = -24 + 12 = -12

Step 3: Find f(2), so you have a point on y=f(x):

f(2) = -2·(2)^3 + 3·(2)^2 = -16 + 12 = -4

So, you have the point (2,-4) and the slope of -12.

Step 4: Find the equation of your tangent line:

Using point-slope form you'd have: y + 4 = -12 (x - 2)

That is the equation of the tangent line.

If your teacher is picky and wants slope-intercept, solve that for y to get:

y = -12 x + 20

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not riding a motor cycle given that they read

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3 years ago
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Rom4ik [11]

Answer: They are parallel

Step-by-step explanation:

If two lines are parallel , then they must have the same slope and if two lines are perpendicular , the product of their slope must be -1.

To check this , we must calculate the slope of the two lines given.

Slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}

from the first point

y_{1} = 2

y_{2} = 1

x_{1} = 5

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substituting the values

slope 1 = 1 - 2 / -3 - 5

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Using the same format to calculate the slope of the second line

y_{1} = -2

y_{2} = 0

x_{1} = -1

x_{2} = 15

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Since slope 1 = slope 2 , this implies that the lines are parallel

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3 years ago
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