Tangent Line Problem: Find Point P & Compute Slope m_L

CrossFit415
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I'm on mobile so I can't use latex.

Let C: y=8x^5+5x+1 and suppose L is a line through the origin tangent to C at a point P=(a,f(a)) on C.

-Find the coordinates of P

-Compute the slope m sub L of L

Where should I begin? I'm guessing I would need the derivative of the equation f(x)? Then I would use the point slope form to figure out point P?
 
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CrossFit415 said:
I'm on mobile so I can't use latex.

Let C: y=8x^5+5x+1 and suppose L is a line through the origin tangent to C at a point P=(a,f(a)) on C.

-Find the coordinates of P

-Compute the slope m sub L of L

Where should I begin?
Maybe a sketch of the function?
CrossFit415 said:
I'm guessing I would need the derivative of the equation f(x)?
One would think that might somehow enter into things.
CrossFit415 said:
Then I would use the point slope form to figure out point P?
 
After I sketch the eqiation what would I do after? So does that mean L is a secant line through the equation?
 
L is a line that is tangent to the graph of the function. It's not a secant line (a line that hits a curve at two points).

BTW, "a secant line through the equation" doesn't make much sense, unless you actually draw a line through an equation, as opposed to drawing a line through the graph of an equation.
 
Mark44 said:
L is a line that is tangent to the graph of the function. It's not a secant line (a line that hits a curve at two points).

BTW, "a secant line through the equation" doesn't make much sense, unless you actually draw a line through an equation, as opposed to drawing a line through the graph of an equation.

Haha I'll be more precise. I misread I thought it was going through the graph of the equation thinking its a secant line.

So I got the derivative which is f'(x) = (40x^4) + 5.

I wouldn't be able to apply the slope formula since I'm figuring out the coordinates of P. What methods are there?
 
CrossFit415 said:
suppose L is a line through the origin tangent to C at a point P=(a,f(a)) on C.

So you have a line from (0, 0) to (a, f(a)) on the graph of your curve. Write an expression that represents the slope of the line.

Write another expression that represents the slope of the tangent at any point on your curve.

Equate the two expressions.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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