Find k for Tangent Line of f(x) = x^2 - kx

[bubbles]

Hi, I need help on a math problem that asks me to find k such that the line is tangent to the function, given$$f(x) = x^2 - kx$$ and $$y = 4x - 9$$.

I don't know how to solve for k. Is "tangent line" the same thing as the equation that is the derivative of the function?

 

[chroot]

To find a tangent line:

1) Find the derivative of your function. The derivative gives you the slope of the function for any given x.

2) Find a point on the curve: an x₀ and its corresponding y₀.

3) Use point-slope form to combine the point and slope into a single equation.

In this case, the x₀ is arbitrary. You're not trying to find a particular tangent line at a particular x₀, you're trying to find the tangent line for any arbitrary x₀.

- Warren

 

[MeJennifer]

Is "tangent line" the same thing as the equation that is the derivative of the function?

No, while the slope of a tangent line at a particular value x for a given function f(x) is the derivative of that function for the particular value x, the derivative of a function f(x) is the expression of the slopes of the tangent lines for all values of x.

 

[bubbles]

How do I find a point of the curve f(x)=x^2-kx where there are 2 variables?

 

[chroot]

Every point on that curve is of the form (x, f(x)). k cannot be eliminated until the very end of this problem.

- Warren

 

[HallsofIvy]

How do I find a point of the curve f(x)=x^2-kx where there are 2 variables?

There aren't. There is only one variable, x (you know that because it says "f(x)"). The k is an unknown constant. What is the derivative of f(x)= x²- kx, remembering that the variable is x? Knowing that x must give the same "y" value for y= x²- kx and y= 4x- 9 gives you one equation for the two unknown numbers x and k. Knowing that the derivative of f(x)= x²- kx at that x is the same as the slope of y= 4x- 9 gives you another. Now you have two equations for the two unknown numbers, x and k.

(There are, by the way, two correct solutions.)