Question 982609
Slope of the tangency line must be same as dy/dx for the parabola, and this dy/dx will be a variable expression. 

This slope based on the parabola:
{{{(dy/dx)(4x-x^2)=4-2x}}}
Again, this is a SLOPE for the as yet unknown tangent lines.


Here is a look just at the parabola:
{{{graph(300,300,-6,6,-6,6,4x-x^2)}}}
You should be able to understand that (2,5) is on the axis of symmetry and is 1 unit above the vertex.


The tangent line built using Point-Slope equation form:
This must contain the given point (2,5);
{{{y-5=(4-2x)(x-2)}}}
and doing the algebra steps....
{{{highlight_green(y=-2x^2+8x-3)}}}----Understand this without its appearance fooling you.  This represents a LINE, but it is variable.


If you sketch the graph, the understanding should intuitively be this:
We must have  (tangent line) - (parabola) = 0, in order for the intersection to be only any single point.
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Use the formulas of y for each of those objects.
{{{-2x^2+8x-3-(4x-x^2)=0}}}
Do the algebra steps...
omitting them here...
{{{highlight_green(highlight_green(x^2-4x+3=0))}}}----------This will give the x coordinates of the two tangent lines at the points of tangency on the parabola.
... and it is FACTORABLE.


x=1  or  x=3


Use each individually in the line equation  {{{y-5=(4-2x)(x-2)}}}, which is equation for line intersection the parabola.  ONLY substitute for the slope, which is  {{{4-2x}}}; so that you can get the equation of the TANGENT line.
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Doing that and simplifying :
{{{y=(4-2*1)(x-2)+5}}}
{{{highlight(y=2x+1)}}}
and
{{{y=(4-2*3)(x-2)+5}}}
{{{highlight(y=-2x+9)}}}
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The two tangent lines.