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#### alvintran12

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These are my steps (Please do the problem, but also tell me where my mistake was):

Step 1: Deriv of curve
y'=3x^2

Step 2: Slope of Parallel line
m=12

Step 3: Plug 12 into y' for slope
12=3x^2
x= 2,-2

Step 4: Plug 2 into Original Equation to get points.
y(2)=1+8
y(2)=9
y(-2)=1-8
y(-2)=-7
Points= (2,9);(-2,-7)

Step 5: Plug points into y=mx+B equation with m=12 to get the equation of the lines tangent to y=1+x^3 but parallel to 12x-y
9=12(2)-1+b
9=24-1+b
b=-14
-7=12(-2)+b
b=-31

Step 6: Write equations of the lines because I found b.
y=12x-14 and y=12x-31

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nvm

#### alvintran12

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The back of the book says this is the answer:

#### SirLagALot

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Made a mistake, ignore my previous post
Also, image doesn't work
Again, mistake is in step 5

Quote:
 Originally Posted by alvintran12 Step 5: Plug points into y=mx+B equation with m=12 to get the equation of the lines tangent to y=1+x^3 but parallel to 12x-y 9=12(2)-1+b 9=24-1+b b=-14 -7=12(-2)+b b=-31 Step 6: Write equations of the lines because I found b. y=12x-14 and y=12x-31
So you want the equations of the line in the form y = mx + b
You don't actually need that -1
Ends up with b = -15

You messed up the signs here
So you have -7 = -24 + b
You end up with 17 = b

#### alvintran12

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Nvm we got it!!! YAY You rock!!!

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