Question 1201007
<pre>
The equation of a line through a point (x<sub>1</sub>,y<sub>1</sub>)
with slope m is {{{y-y[1]=m(x-x[1])}}}

The equation of every line through (4,-1) with slope m is

{{{y-(-1)=m(x-(4)^"")}}}
{{{y+1=mx-4m}}}
{{{y=mx-4m-1}}}

Usually if a line intersects a parabola, it intersects it in TWO points
However, a tangent line must intersect a parabola in only ONE (not TWO!)
points.  So when we solve the system of the parabola and the line:

{{{system(y=x^2+4x+3,y=mx-4m-1)}}}

we must get only ONE solution.  We set the right sides equal:

{{{x^2+4x+3=mx-4m-1}}}
{{{x^2+4x-mx+4+4m+4=0}}}
{{{x^2+(4-m)x+(4m+4)=0}}}

To guarantee this has only ONE solution, the discriminant must be 0.

{{{Discriminant}}}{{{""=""}}}{{{b^2-4ac}}}, a=1, b=4-m, c=4m+4.

{{{Discriminant}}}{{{""=""}}}{{{(4-m)^2-4*1*(4m+4)}}}{{{""=""}}}{{{16-8m+m^2-16m-16}}}{{{""=""}}}{{{m^2-24m}}}
So we set this discriminant equal to 0:

{{{m^2-24m=0}}}
{{{m(m-24)=0}}}
m=0; m-24=0
       m=24
So the equations of the tangent lines are

{{{y=mx-4m-1}}} with m=0, and m=24

{{{y=0x-4(0)-1}}} and {{{y=24x-4(24)-1}}}

{{{y=-1}}} and {{{y=24x-97}}}   <---answers

The two green lines are the tangent lines.  The points of tangency are at the vertex of the parabola (-2,-1) and the point (10,143)

{{{drawing(400,1000,-12,12,-19,150,graph(400,1000,-12,12,-19,150,169,-1),
graph(400,1000,-12,12,-19,150,x^2+4x+3,24x-97))}}}

Edwin</pre>