Question

I know how to find the eqn of a tangent line given the point of tangency, but that's not what I have. The function f(x)=4x-(x^2) has two lines drawn tangent to it, on either side of the parabola. But, the only coordinates given are (2,5), where the two tangent lines cross each other as they extend from touching the parabola. We get the answer to the question, which is to find the equations of the tangent lines, and the answers are y=2x+1 and y=-2x+9, but I don't know why.
Here's what I did so far:
f'(x)=4-2x
4-2x=(y-5/x-2)=(4x-(x^2)-5)/x-2
but then I couldn't factor the numerator, and it gets very frustrating and confusing, thank you for the help!

Answer 1

you are on the right track
\[ 4-2x=\frac{(4x-x^2-5)}{x-2} \\
(4-2x)(x-2) = -x^2 + 4x - 5 \\
-2x^2 +8x -8 = -x^2 + 4x - 5 \\
0 = x^2 -4x +3 \\
(x-3)(x-1)=0 \\
x= 1, \ x=3
\]

Answer 2

using x=1, you get the point (1,3) on the parabola
for x=3, you get point (3,3)

you can now find the equations of both lines , which share the point (2,5)