#x# intercepts
To find the #x# intercepts there are 3 methods. These methods are factorisation, quadratic formula, and completing the square. Factorising is the easiest method but does not work all the time, however it does in your case.
To factorise the expression we must create two brackets: #(x+-f)(x+-g)# We can figure out the values of a and b from the equation above.
The general form of a quadratic equation is #ax^2 + bx + c#. The values of #f# and #g# must multiply to make #c# which in your case is 4. The values must also and add together to make #b# which in your case is -4. This example is easy, as both #a# and #b# are -2 and this satisfys both conditions above. So our factorised equation is #(x-2)(x-2)#
The solutions to the equation are the opposite value to those in the brackets. In this case this means the solutions are both just 2, and there is only one solution so there is only one point where it crosses the #x# axis. Note that in examples where the brackets have a different value in them then there will be 2 points where the line crosses the #x# axis.
To find the #y# coordinate of this point we substitute our value of #x#, 2 into the original equation.
#y = (2)^2 - 4(2) + 4#
#y = 4 - 8 + 4#
#y = 0#
So the value of #y# is 0 at this point, and our #x# intercept coordinate is #(2,0)#. If you got two values for #x# in the previous part you would have to do this twice to get both coordinates.
#y# intercept
The #y# intercept is much easier to find. As we know on the #y# intercept the value of #x# is equal to 0. Therefore we just substitute this into the equation to find the value for #y#.
#y = (0)^2 - 4(0) + 4#
Removing everything multiplied by 0 we get: #y = 4#
So therefore the #y# intercept coordinate is #(0,4)#.
