Question 992501
<pre>
A tangent line can be thought of as a line that "intersects
a curve twice at the same point".

The value of y when x = 2 is

{{{y = -x^2 + 3x + 8}}}
{{{y = -(2)^2 + 3(2) + 8}}}
{{{y = -4 + 6 + 8}}}
{{{y = 10}}}

So we want the equation of a tangent line at the point (2,10)

That is, we want to find the equation of the line through (2,10)
that is tangent to the parabola whose is equation is 
{{{y = -x^2 + 3x + 8}}}

{{{drawing(3600/17,400,-3,6,-3,14,

graph(3600/17,400,-3,6,-3,14,-x^2 + 3x + 8),

circle(2,10,0.15),circle(2,10,0.13),circle(2,10,0.11),circle(2,10,0.09),circle(2,10,0.07),circle(2,10,0.05),circle(2,10,0.03),circle(2,10,0.01),
locate(2.3,10.5,"(2,10)") )}}}

Let the tangent line have the equation 

{{{y = mx+b}}}

The given equation of the parabola is

{{{y = -x^2 + 3x + 8}}}

We set the two expressions for y equal to each other:

{{{mx+b = -x^2+3x+8}}}
{{{x^2+mx-3x+b-8 = 0}}}
{{{x^2+(m-3)x+(b-8)}}}

That quadratic will have a double solution if its
discriminant is 0

The discriminant is {{{B^2-4AC}}}.  We use capital
letters to distinguish little b and capital B.

{{{discriminant = B^2-4AC=(m-3)^2-4(1)(b-8) = (m-3)^2-4(b-8)}}}

{{{(m-3)^2-4(b-8)=0}}}

Solve that for b

{{{m^2-6m+9-4b+32=0}}}
{{{m^2-6m-4b+41=0}}}
{{{m^2-6m+41=4b}}}
{{{expr(1/4)(m^2-6m+41)=b}}}

Substitute that for b in

{{{y = mx+b}}}

{{{y = mx+expr(1/4)(m^2-6m+41)}}}

And also the point (x,y) = (2,10)

{{{10 = m(2)+expr(1/4)(m^2-6m+41)}}}

{{{10 = 2m+expr(1/4)(m^2-6m+41)}}}

Multiply through by 4 to clear fraction:

{{{40 = 8m+(m^2-6m+41)}}}

{{{40 = 8m+m^2-6m+41}}}

{{{0 = m^2+2m+1)}}}

{{{0 = (m+1)^2}}}

{{{m+1=0}}}

{{{m=-1}}}

Substitute this and (x,y) = (2,10) in

{{{y = mx+b}}}

{{{10 = (-1)(2)+b}}}

{{{10 = -2+b}}}

{{{12=b}}}

So the equation of the line is 

{{{y = -x+12}}}

So we draw that line on the graph to see if it looks like what we
have calculated. It has y-intercept and slope -1, and goes through
points (-1,13), (0,12), (4,8)

{{{drawing(3600/17,400,-3,6,-3,14,

graph(3600/17,400,-3,6,-3,14,-x^2 + 3x + 8),
line(-8,20,7,5),
circle(2,10,0.15),circle(2,10,0.13),circle(2,10,0.11),circle(2,10,0.09),circle(2,10,0.07),circle(2,10,0.05),circle(2,10,0.03),circle(2,10,0.01),
locate(2.3,10.5,"(2,10)") )}}}

Edwin</pre>