Consider the following function. Complete parts (a) through (e) below. \( y=x^{2}+4 x+3 \) The parabola opens downward. The parabola opens upward. b. Find the vertex. The vertex is (Type an ordered pair.) c. Find the \( x \)-intercepts. The \( x \)-intercepts are (Type an integer or a fraction. Use a comma to separate answers as needed.) d. Find the \( y \)-intercept. The \( y \)-intercept is \( \square \). (Type an integer or a fraction.) e. Use the results from parts (a)-(d) to graph the quadratic function.

The parabola represented by the function \( y = x^2 + 4x + 3 \) opens upward, as indicated by the positive coefficient of the \( x^2 \) term. This means you'll see that beautiful U-shape when you plot it! Next, let’s calculate the vertex. The vertex form for a parabola in the form \( y = ax^2 + bx + c \) is found using the formula \( x = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = 4 \), so the x-coordinate of the vertex is \( x = -\frac{4}{2 \cdot 1} = -2 \). Plugging this back into the original equation gives us \( y = (-2)^2 + 4(-2) + 3 = -1 \). Thus, the vertex is at the point \((-2, -1)\). For the \( x \)-intercepts, we need to set \( y = 0 \): \[ 0 = x^2 + 4x + 3 \] Factoring gives us \((x + 1)(x + 3) = 0\), yielding intercepts at \( x = -1 \) and \( x = -3\). Therefore, the \( x \)-intercepts are \(-1, -3\). Now, let’s find the \( y \)-intercept by plugging in \( x = 0 \): \[ y = (0)^2 + 4(0) + 3 = 3. \] So, the \( y \)-intercept is \( 3 \). To graph the quadratic, plot the vertex at \((-2, -1)\), the \( x \)-intercepts at \((-1, 0)\) and \((-3, 0)\), and the \( y \)-intercept at \((0, 3)\). Connect these points with a smooth curve, and voila! You have your beautiful parabola!