Question
Asked Oct 20, 2019

Find equations of both lines that are tangent to the curve y = x3 - 3x2 + 3x - 3 and are parallel to the line 3x - y = 15.

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Step 1

The given curve is represented by y = x3 - 3x2 + 3x - 3 and are parallel to the line 3x - y = 15.

Slope of line 3x-y=15 is given by,

3х — у %3D15
Зх.
-3х +15
у%3 3х -15
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3х — у %3D15 Зх. -3х +15 у%3 3х -15

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Step 2

Since the slope of the parallel lines are equal so, the slope of the tangent line will be equal to 3.

Now differentiate the given curve with respect to x in order to find the slope using the curve equation.

d
(x3- 3x2+3x - 3)
dx
(у
dx
3x2-6x+3
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d (x3- 3x2+3x - 3) dx (у dx 3x2-6x+3

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Step 3

Now equate both the slope...

3 3x2-6x3
3x2-6x 0
x2-2x 0
x(x-2)0
x = 0 or x = 2
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3 3x2-6x3 3x2-6x 0 x2-2x 0 x(x-2)0 x = 0 or x = 2

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Tagged in

Math

Calculus

Derivative

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