The intersection of the line y=mx-5 and quadratic graph can be investigated below. For two special m values the line
touches the quadratic graph i.e. there is only one intersection point; can you find these special m values? .

change line gradient m

The Analytic Solution

The special line gradients for which the line touches the parabola can be found analytically i.e. no need to
experiment by plotting many lines. You may have noticed that when the discriminant=0 there is only one solution;
and in this problem only one intersection point.

The line has gradient m :

y=mx-5

At the intersection point(s):

x^{2}-8x+19=mx-5

rearranging the equation:

x^{2}-(m+8)x+24=0

The above equation is then compared with the general form of the quadratic equation
y=ax^{2}+bx+c
we find that, a=1, b=-(m+8) , and c=24 .
When the discriminant b^{2}-4ac=0 there is only one solution i.e. only
one intersection point, and the line touches the quadratic equation. The discriminant is a function of m and we solve
m this is unusual but it is correct. We need find the m value so that b^{2}-4ac=0 :

(m+8)^{2}-96=0

add 96 to both sides

(m+8)^{2}=96

taking the square root of both sides:

m+8=±√(96)

m=-8±√(96)

The two m values rounded to 3 decimal places are then m=1.798 and m=-17.798 .

Conclusion

When the discrimant b^{2}-4ac>0 the quadratic equation has two solutions. When the discriminant
b^{2}-4ac<0
the quadratic equation has no solutions, and when the discriminant
b^{2}-4ac=0 , there is only one solution.