tangent of a line?

[Failenn]

I have to find k such that the line is tangent to the graph of the function.

given f(x) = x^2 - kx. with line given y = 4x - 9

my reasoning is that i need to take the derivative first.... and so i get

f'(x) = 2x - k

after this I'm not sure how to proceed or if I should even be finding the derivative first.

 

[Deleted member 4993]

I have to find k such that the line is tangent to the graph of the function.

given f(x) = x^2 - kx. with line given y = 4x - 9

my reasoning is that i need to take the derivative first.... and so i get

f'(x) = 2x - k

after this I'm not sure how to proceed or if I should even be finding the derivative first.

Your first step is correct - you need to find the derivative.

Now the question is - how is the derivative of a curve is related to the tangent to the curve?

What does your textbook/class-notes say about this?

 

[wjm11]

I have to find k such that the line is tangent to the graph of the function.

given f(x) = x^2 - kx. with line given y = 4x - 9

my reasoning is that i need to take the derivative first.... and so i get

f'(x) = 2x - k

Consider this: you are starting off with two equations, y = x^2 – kx and y = 4x – 9, but there are three unknowns; x, y, and k. By finding the derivative, we can create a third equation. Set the derivative equal to the slope of the line:

2x – k = 4

You now have three equations and three unknowns, sufficient information to find the solution.

 

[BigGlenntheHeavy]

\(\displaystyle y \ = \ x^{2}+kx, \ \implies \ y \ ' \ = \ 2x+k\)

\(\displaystyle y \ = \ 4x-9, \ m \ =4\)

\(\displaystyle 2x+k \ = \ 4, \ \implies \ k \ = \ 4-2x\)

\(\displaystyle y \ = \ x^{2}+(4-2x)x \ = \ 4x-x^{2}\)

\(\displaystyle 4x-x^{2} \ = \ 4x-9 \ \implies \ x \ = \ \pm3\)

\(\displaystyle Ergo, \ when \ x \ =3, \ k \ =-2 \ and \ when \ x \ = \ -3, \ k \ = \ 10\)

\(\displaystyle Hence, \ y \ = \ x^{2}-2x, \ y \ = \ x^{2}+10x, \ and \ y \ = \ 4x-9, \ see \ graph \ below.\)

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