Finding equation of tangent y = x^4 + 1 that is parallel to the line 32x - y = 15

[ricecrispie]

Hi again,

I'm not sure I'm approaching this problem correctly and there's no solution at the back of my textbook so I'm hoping to get some feedback from here.

Q:

Find an equation of the tangent line to the curve y = x^4 + 1 that is parallel to the line 32x - y = 15

Firstly:

y' = 4x^3 = m
The gradient from the 2nd function is 32

So now 4x^3 should = 32, as parallel lines have the same gradient.

4x^3 = 32
x^3 = 8
x = 2

Now we know that y'(2) = 32

To get a our co-ordinates we can use y:
y = 2^4 + 1 = 17 ......... (2, 17)

Now the tangent line will be:
y - 17 = 32(x-2)
y = 32x - 47

Is this correct?


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[mmm4444bot]

… there's no solution at the back of my textbook …

y = 32x - 47

Is this correct?

Looks good, to me.

One way to (visually) confirm this result is to graph all three functions.

 

[ricecrispie]

Yes, the lines looked parallel to me. I just wasn't sure if my method is correct. Thanks for the feedback !

Looks good, to me.

Did you graph all three functions, in order to visually check your result?

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[mmm4444bot]

… I just wasn't sure if my method is correct …

Unless you're instructed to use a specific method, any method you choose that leads to the solution is correct.

Good job. :cool: