Graphing Quadratic Functions I need help with $f(x)=(x-3)^2$. My teacher said the $x$ intercept is $3$. I don't understand how it   can be positive $3$ when it is negative $3$ in the parenthesis.    
 A: The $x$-intercepts are the points where $f(x)=0$, i.e., where the graph touches the $x$-axis. You have $f(3)=(3-3)^2=0$, so you have an $x$-intercept at $x=3$.
A: Lets graph it and see what happens.

Do you see why $f(x) = 0$ at $x=3$ now?
A: $x$-interscepts are when the function intersects the $x$ axis, i.e. when $f(x)=0$.  Thus
$$0=(x-3)^2 \implies 0=x-3\implies {x=3}$$
A: The x-intercept is the value that you can plug in for x and have f(x) come out to 0. Try evaluating f(3), and see what happens.
Since f(3)=0, that means the point (3,0) is on the graph, so it is an x-intercept.
Seeing the minus sign inside the parentheses can be confusing, because it makes you think of a negative number, but remember to ask, "what value of x would make this equal 0?"
A: The $x$ intercepts are the abscissa of intersection points of the curve with the $x$ axis. When the curve intersects the $x$ axis, the ordinate is always equal to 0. 
 
The code to generate the graph can be found here.
Therefore, we have the following equation:
$$
(x-3)^2=0
$$
We have a theorem as follows.

The solution of  $$ A \times B =0 $$
is $A=0$ with any $B$ OR any $A$ with $B=0$.

Thus, the solution of $(x-3)^2=(x-3)(x-3)=$ is $x-3=0$ or $x=3$.
