(03.03 MC) Given f(x)=(x-1)(x+2)(x-3) what are the zeros and end behavior of the function? -1 2. -3 continues downward to the left and upward to the right -1 2. -3 continues upward to the left and downward to the right 1,-2,3 continues downward to the left and upward to the right 1,-2,3 continues upward to the left and downward to the right

## Step 1: Finding the Zeros<br />### To find the zeros of the function $f(x) = (x-1)(x+2)(x-3)$, set $f(x)$ equal to zero and solve for $x$. This means setting each factor equal to zero: $x - 1 = 0$, $x + 2 = 0$, and $x - 3 = 0$.<br /><br />## Step 2: Solving for x<br />### Solving each equation gives us the zeros: $x - 1 = 0 \Rightarrow x = 1$, $x + 2 = 0 \Rightarrow x = -2$, and $x - 3 = 0 \Rightarrow x = 3$. Therefore, the zeros are 1, -2, and 3.<br /><br />## Step 3: Determining End Behavior<br />### The degree of the polynomial is 3 (since it's a product of three linear factors), which is odd. The leading coefficient is positive (since expanding the expression would result in a positive $x^3$ term). An odd degree with a positive leading coefficient means the graph goes down to the left and up to the right.