Find k so that the line y = 3x-4 is tangent to the graph of f(x) = x^2 + kx.

Question

I got k = 7, -1 using substitution on  2x+k = 3 and x^2 + kx = 3x - 4.

Is this correct?

Answer 1

Yes, you are correct.

 

 

For equating the equations, you get

 

x² + kx - 3x + 4 = 0               eq1

 

 

Setting the derivative of f(x) equal to 3 yields

 

f'(x) = 3

 

2x + k = 3                           eq2

 

 

Now we substitute the value of k from eq2 into eq1.

 

x² + x(3 - 2x) - 3x + 4 = 0

 

 

Solve for x from this equation.  Then plug in the value of x into eq2 to solve for k