Consider the parabola $$ y=(3-x)(x+1) $$. Find the $$ y $$ value of the $$ y $$ - intercept. $$ \quad y=3 $$ Find the $$ x $$ values of the $$ x $$-intercepts Write all solutions on the same line separated by a comma.

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To find the $$ y $$ value of the $$ y $$-intercept, we need to set $$ x = 0 $$ in the equation of the parabola and solve for $$ y $$.

Given equation: $$ y = (3-x)(x+1) $$

Setting $$ x = 0 $$:
$$ y = (3-0)(0+1) $$
$$ y = 3 $$

Therefore, the $$ y $$ value of the $$ y $$-intercept is $$ y = 3 $$.

To find the $$ x $$ values of the $$ x $$-intercepts, we need to set $$ y = 0 $$ in the equation of the parabola and solve for $$ x $$.

Setting $$ y = 0 $$:
$$ 0 = (3-x)(x+1) $$

Expanding the equation:
$$ 0 = 3x + 3 - x^2 - x $$
$$ 0 = -x^2 + 2x + 3 $$

Now, we need to solve the quadratic equation $$ -x^2 + 2x + 3 = 0 $$ to find the $$ x $$ values of the $$ x $$-intercepts.
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
  $$-x^{2}+2x+3=0$$
- step1: Factor the expression:
  $$\left(-x+3ight)\left(x+1ight)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&-x+3=0\&x+1=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&x=3\&x=-1\end{align}$$
- step4: Rewrite:
  $$x_{1}=-1,x_{2}=3$$
The solutions to the quadratic equation $$ -x^2 + 2x + 3 = 0 $$ are $$ x = -1 $$ and $$ x = 3 $$.

Therefore, the $$ x $$ values of the $$ x $$-intercepts are $$ x = -1 $$ and $$ x = 3 $$.

The solutions are $$ x = -1, 3 $$.
The $$ y $$-intercept is $$ y = 3 $$, and the $$ x $$-intercepts are $$ x = -1 $$ and $$ x = 3 $$.

Consider the parabola \( y=(3-x)(x+1) \). Find the \( y \) value of the \( y \) - intercept. \( \quad y=3 \) Find the \( x \) values of the \( x \)-intercepts Write all solutions on the same line separated by a comma.

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