Let f(x)=x^2-ax+a-1 and g(x)=x-5. The graphs of f and g intersect at one distinct point. Find the possible values of a.

If the two graphs intersect, it means that they will share the same y and x coordinates at one particular point. (I will draw diagram to show point).

Therefore, you can set f(x)=g(x) so that x^2 -ax +a -1 = x-5 Then, x^2 -x(a+1) +a + 4 = 0

If they only intersect at one particular point, this means that the previous quadratic equation has only one solution. This is translated into an equation in terms of the determinant so that the determinant must be 0.

(If necessary, I will explain the difference number of solutions that one gets for different determinants).

Then, one requires b^2 - 4ac = 0 , where b is the coefficient multiplying the x, a is the coefficient multiplying the x^2 and c is the coefficient with no x in the previous equation.

This leads to (a+1)^2-4(a+4)=0 which is a quadratic equation in a:

a^2 -2a -15 =0 , which, using the quadratic formula, has solutions a= 5 and a=-3.