Let's create a table, and use arbitrary values of x between -3 and 3 (if the solution is not in this interval, we can use another one later). If we call:
[tex]f(x)=x^2-4x+4[/tex]
We want to find the x's for with f(x) = 0
This is the table:
And now we can evaluate f(x) in each value of x to complete the table:
[tex]\begin{gathered} f(-3)=(-3)^2-4\cdot(-3)+4=9+12+4=25 \\ f(-2)=(-2)^2-4\cdot(-2)+4=4+8+4=16 \\ f(-1)=(-1)^2-4\cdot(-1)+4=1+4+4=9 \\ f(0)=0^2-4\cdot0+4=4 \\ f(1)=1^2-4\cdot1+4=1-4+4=1 \\ f(2)=2^2-4\cdot2+4=4-8+4=0 \\ f(3)=3^2-4\cdot3+4=9-12+4=1 \end{gathered}[/tex]
The table is:
If we use this values and plot them in the cartesian plane.
We get:
And now if we plot a line that connects the points, we get the graph of the quadratic equation:
Since the problem ask us to find the value of x for which f(x) = 0, we can see both in the table and in the graph that this value is x = 2
