How to Find Two Parallel Tangents for a Cubic Function?

[jahaddow]

Homework Statement

I have a cubic function, y=(x-6)(x-1)(x-9) or y=x³-16x²+69x-54
I then have two tangents, y=7.75x+14.75 and y=-6.25x+56.25
What I need to find is another two tangents that are parallel.

Homework Equations

The Attempt at a Solution

What I know needs to be found is the two x points which the tangents cross, the rest from there on is simple enough. So how do I find the two x points parallel to the other tangents. Any Help is much appriectated.

 

[Office_Shredder]

To find a tangent that is parallel, you need to find one with the same slope. How can you find a tangent line with slope 7.75?

 

[jahaddow]

Sorry, I'm really blank about the whole thing, Please explain

 

[Mark44]

One of your tangent lines has a slope of 7.75. This means that at some point on the graph of the cubic, the value of the derivative function is 7.75. Find the derivative of your cubic function and find both values of x for which dy/dx = 7.75.

Do the same thing for the other tangent line, the one whose slope is -6.25.

 

[jahaddow]

I really don't know how to do that

 

[Mark44]

Your book should have some examples showing how to find the derivative of a given function. The derivative can be used to find the slope of a tangent line at any point on the graph of the function whose derivative you take.

For example, if f(x) = 2x² + 3x, f'(x) = 4x + 3. At x = 0, f'(0) = 4(0) + 3 = 3. At (0, 0), the slope of the tangent line is 3.

 

[jahaddow]

can you tell me the answer to one of the tangents, and then I will work out the other myself?

 

[jahaddow]

I'm not trying to find the y value, just the x value, the y value is easy to find after getting the x value.

 

[Mark44]

can you tell me the answer to one of the tangents, and then I will work out the other myself?

No, that goes against the rules in this forum. You should be able to find the derivative of y = x³ - 16x² + 69x-54, and set the derivative to 7.75 and -6.25, respectively, to find all of the points of tangency for those values.

You need to make an effort at this yourself. I'm willing to help, but I'm not willing to do the work for you.

 

[jahaddow]

3x^2-32x+69 then what?

 

[novop]

You want the x (or y) coordinate of the points at which the slope of your function is 7.75. To do this, like Mark said, set your derivative function equal to 7.75.

f'(x)=3x^2-32x+69=7.75

Now solve that equation. Use the quadratic formula. You will get two values for x, one you have already, and the second being the one you're looking for.

 

[Mark44]

And do the same thing for the other slope, setting the derivative to -6.25.
3x^2 - 32x + 69= -6.25