Both graph intersect when f1=f2

==> 2x^2 - 5x + 10 = x^2 -11x + 1

Now group similar:

==> x^2 +6x +9 = 0

==> (x+3)^2 = 0

The equatio has one solution:

==> x=-3

Now to determine y value , substitute with f:

==> y = -3^3 -11(-3) +1

==> y= 9+33+1

==> y= 43

Then both function has one common point at (-3,43)

At the common point f(x) = 2x^2-5x+10 and f(x) = x^2-11x+1 both curves have the same ordinate f(x) .So

2x^2-5x+10 = x^2 -11x+1 .

2x^2-5x+10 -(x^2-11x+1) = 0

2x^2-x^2 +(-5x+11x)+10-1 = 0

x^2+6x+9 = 0

(x+3)^2 = 0

x +3 = 0 or x = -3 .

At x=-3, from f(x) = x^2-11x+1 f(-3) = (-3)^2-11(-x) +1 = 9+33+1 = 43.

So (-3,43) is the coordinates of the common point

To prove that the graphs are intercepting in a single point, we have to solve the system formed from the equations of the given functions and the system has to have just one solution.

We could write the system in this way:

2x^2-5x+10=x^2-11x+1

We'll move all terms to one side and we'll combine like terms:

2x^2-5x+10-x^2+11x-1 = 0

x^2 + 6x + 9 = 0

For this equation to have just one solution, it's discriminant has to be zero.

delta=0

delta=

(6)^2-4*9 = 0

36-36 = 0

x1 = x2 = 6/2

x1 = x2 = 3

y1 = y2 = 3^2-11*3+1 = 9-33+1 = -23

The system has just one solution, so the graphs are intercepting each other in a single point.

The coordinates of the intercepting point are: (3 , -23).