I need help understanding how to solve this, my teacher barely introduced it to us today in class but it was in the homework and we didn’t work through enough problems for me to understand

Question

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

Answer 1

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

A tangent can only occur when the two lines have a common point.

so equating both equation

3x -4 = x² + kx

x² +kx -3x +4 =0

x² + x(k-3) + 4 =0 __________________ eq 1

And at that point they have the same slope

so

slope of line y= 3x- 4 is 3

slope of function f(x) =x² +kx is

f'(x) = 2x + k

so 2x + k = 3

k =3-2x _______________eq 2

substituting k=3 -2x in eq1

x² + x(k-3) + 4 =0

x² + x(3-2x-3)+4 =0

x² + x(-2x)+4 =0

x² - 2x² + 4 =0

-x² +4 =0

x² = 4

x =±2

so

k =3-2x

= 3-2(2) =3-4=-1 OR 3-2(-2) =3+4=7

So k = 7 or -1