## Find k such that the line is tangent to the function.

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### Find k such that the line is tangent to the function.

I am having extreme trouble with this differentiation problem. It says, "Find k such that the line is tangent to the graph of the function."

Function
F(x) = x^2 - kx

Line
y = 4x - 9

I don't understand the step where you set f(x) to equal y. How do you get rid of the k?
LastXdeth
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### Re: Fine k such that the line is tangent to the function.

$\frac{d}{dx}\,(x^2\,-\,kx)\,=\,4$ so the slope of the line and the function are the same (one condition of tangency).

$x^2\,-\,kx\,=\,4x\,-\,9$ so the line and curve intersect (other condition of tangency).

$\frac{d}{dx}\,(x^2\,-\,kx)\,=\,2x\,-\,k\,=\,4,\,k\,=\,2x\,-\,4$

$x^2\,-\,(2x\,-\,4)x\,=\,4x\,-\,9$

$x^2\,-\,9\,=\,0\,\Rightarrow\,x\,=\,\pm\,3$

so k = 2, -10.

greg1313
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### Re: Fine k such that the line is tangent to the function.

Fermat, prior to Newton and Leibniz, had a method for problems like this. It uses the fact that a tangent line has a "second order" intersection with a curve. That is, if f(x) is tangent to t(x) at $x_0$, $x_0$ must be a double root of f(x)= g(x). Your given curve is $y= x^2- kx$ and your desired tangent line is $y= 4x- 9$ so we look at the equation $x^2- kx= 4x- 9$ which is the same as $x^2- (k+4)x+ 9= 0$.

By the quadratic formula, solutions are given by $x= \frac{k+4\pm\sqrt{(k+4)^2- 36}}{2}$. There will be a double root if and only if the discriminant, $(k+ 4)^2- 36= k^2+ 8x- 20= 0$. Solve that equation for two values of k.
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