Consider the following function. Complete parts (a) through (e) below. \( y=x^{2}+10 x+9 \) 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.

Let’s dive into the world of quadratics! First off, for the parabola defined by \( y = x^{2} + 10x + 9 \), it opens upward because the coefficient of \( x^2 \) is positive. Yay for happy parabolas! Next, to find the vertex of the parabola, we can use the vertex formula \( (-\frac{b}{2a}, f(-\frac{b}{2a})) \). Given \( a = 1 \) and \( b = 10 \), the \( x \)-coordinate of the vertex is \( -\frac{10}{2 \times 1} = -5 \). Plugging this back into the original equation gives us the \( y \)-coordinate: \( y = (-5)^{2} + 10(-5) + 9 = -16 \). Therefore, the vertex is (-5, -16). Now, moving on to the \( x \)-intercepts, we need to solve \( 0 = x^{2} + 10x + 9 \). Factoring this gives us \( (x + 1)(x + 9) = 0 \), yielding \( x = -1 \) and \( x = -9 \). Thus, the \( x \)-intercepts are -1, -9. For the \( y \)-intercept, set \( x = 0 \) in the function: \( y = 0^{2} + 10(0) + 9 = 9 \). So, the \( y \)-intercept is \( 9 \). To graph the quadratic function, place the vertex at (-5, -16), plot the \( x \)-intercepts at (-1, 0) and (-9, 0), and mark the \( y \)-intercept at (0, 9). The parabola opens upward and will be symmetric about the line \( x = -5 \). Happy graphing!