Derivative involving an unknown constant

[Arnoldjavs3]

Homework Statement

Determine the value(s) of k such that y=-5k is the equation of the tangent line on the graph of F(x) = -x^2 + 4kx + 1

Homework Equations

n/a

The Attempt at a Solution

not sure where to start this problem; but i understand some fundamentals here. I believe that the tangent will intersect the curve at one point only so I believe that you can equate the two to attempt to solve for k?
Attempt 1:
1. My derivative of the function ended out to be f'(x) = -2(x -2k)
2. I tried implementing this into the original function but I ended up with 4k^2 +1 ( then k = 1/2? )
Attempt 2:
1. After equating the two equations i get -5k = -x^2 + 4kx + 1
2. 0 = -x² + 2.5x + 1 -> x = -0.5 and/or -2
3. I noticed k and x are the same from these two attempts?

What do i do from here? Is everything I've done absolutely garbage or am i on to the correct train of thought?

 

[Stephen Tashi]

2. I tried implementing this into the original function but I ended up with 4k^2 +1 ( then k = 1/2? )

Are you saying that you computed f( f'(x)) ? That wouldn't be relevant to this problem.

You know two things about the tangent line. 1) It goes through some point (x,f(x)) on the graph. 2) The slope of the tangent line is equal to f'(x) at that point. Each of these facts gives you an equation that must be satisfied.

What is the slope of the tangent line given by y = 5k ? (It's a horizontal line.)

 

[Arnoldjavs3]

Okay first i want to thank you for replying; So you're saying that the f'(x) will give me my slope, that I know; (y=5k means the slope is 0)
So in the end..
1. Since my derivative is 0 = -2(x-2k) -> 2k = x? Since hte derivative of f(x) will give me my slope(m) and in this game m is 0, k = x/2? (I actually did this at some point but I didn't know where to continue so I brushed it off)
My question now is that would i simply equate y=-5k in the original function?
Thus further giving me k to be -1/4 and -1.

 

[Arnoldjavs3]

Sorry i forgot to add. I actually did this at one point but my question was I tried confirming if this were the case, but I didn't know what points to input when I got the derivative of the function. To further visualize this:

1. After discovering that k = -1 and -0.25, I found the derivative of the function after inputting k (the function was now -x² -4x + 1, so my derivative ended up being -2(x+2) = m) What would I use as a value of X? I found earlier by subsituting values of X and K i got -2 and -1/2. So in this situation, would I use -2 as the value of X because it would return a value of m = 0?
Sorry if I am unable to verbalize this properly.

 

[Stephen Tashi]

The work would be better organized if you establish the equations that reflect facts 1) and 2). You wrote the equation for fact 2). Write the equation that says the tangent line must intersect the graph of the function.

 

[Arnoldjavs3]

Okay, thank you again. But i just want to confirm; Are these the correct values for K?

Edit: I do believe they are, but I just want to absolutely make sure that I did not make any mistakes

 

[Stephen Tashi]

Okay, thank you again. But i just want to confirm; Are these the correct values for K?

Check if you answers statisfy facts 1) and 2)

You used fact 2) correctly to get the equation x = 2k. What you did after that doesn't make sense to me. ( Just by looking at the facts, one solution to the problem is k = 0, x = 0 )

 

[vela]

You used fact 2) correctly to get the equation x = 2k. What you did after that doesn't make sense to me. ( Just by looking at the facts, one solution to the problem is k = 0, x = 0 )

k=0, x=0 isn't a solution. When k=0, you'd have F(x)=1 and y=0. Those two curves don't intersect.

 

[Stephen Tashi]

k=0, x=0 isn't a solution. When k=0, you'd have F(x)=1 and y=0. Those two curves don't inters ect.

You are correct. I forgot about the +1 in F().