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How do you find the equations of the tangents

  • Thread starter LD745
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  • #1
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How do you find the equations of the tangents to the graph of y=4x^3 that have slope 3?
 

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  • #2
Hurkyl
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It seems to me there are two obvious things to try:
  • Find the points where the slope of the tangent would be 3, then find the lines
  • Find the equations for all tangent lines, set the slope to 3, and solve
Surely you've already had some thoughts on the problem? It's hard to help you effectively if you don't share what you've done and at what points you're stuck.

(It's also against site policy to help people who just post their homework question and don't show any effort of their own...)
 
  • #3
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Oh i have...i found the derivative of 4x^3 which is 12x^2. and i also know that the point-slope formula has to be used.
 
  • #4
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Oh i have...i found the derivative of 4x^3 which is 12x^2. and i also know that the point-slope formula has to be used.
So, in the language of calculus, you have:

[itex]y-y_{1}=\frac{dy}{dx}(x-x_1)[/itex]

[itex]\{x\in \mathbb{R} :4x^{2}=0\}[/itex]

Do you see how you might solve the problem?
 
  • #5
HallsofIvy
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Since this has nothing to do with differential equations, I am moving it to Calculus and Beyond Homework.
 

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