Consider the following function. Complete parts (a) through (e) below. \( y=-2 x^{2}+8 x+11 \) The parabola opens upward. The parabola opens downward. b. Find the vertex. The vertex is (Type an ordered pair.) c. Use the quadratic formula to find the \( x \)-intercepts. The \( x \)-intercept(s) is/are \( \square \). (Type an integer or a decimal rounded to the nearest tenth as needed. 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 a ....-.

The function given is \( y = -2x^2 + 8x + 11 \), which means the parabola opens downward since the coefficient of \( x^2 \) is negative. To find the vertex, we can use the vertex formula \( x = -\frac{b}{2a} \), where \( a = -2 \) and \( b = 8 \). Plugging in these values, we get \( x = -\frac{8}{2 \cdot -2} = 2 \). Substituting \( x = 2 \) back into the function, we find \( y = -2(2^2) + 8(2) + 11 = 15 \). Therefore, the vertex is \( (2, 15) \). Now for the \( x \)-intercepts, we set \( y = 0 \) and solve \( -2x^2 + 8x + 11 = 0 \) using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, we have \( a = -2 \), \( b = 8 \), and \( c = 11 \). Plugging these into the formula gives us \( x = \frac{-8 \pm \sqrt{8^2 - 4(-2)(11)}}{2(-2)} = \frac{-8 \pm \sqrt{64 + 88}}{-4} = \frac{-8 \pm \sqrt{152}}{-4} = \frac{-8 \pm 12.33}{-4} \). This results in \( x \)-intercepts of approximately \( 1.1 \) and \( -2.1 \) (rounded to the nearest tenth). For the \( y \)-intercept, we check where the curve crosses the \( y \)-axis by setting \( x = 0 \). Substituting into the function gives us \( y = 11 \). So, the \( y \)-intercept is \( 11 \). To graph the quadratic, you would plot your vertex at \( (2, 15) \), the \( x \)-intercepts at about \( 1.1 \) and \( -2.1 \), and the \( y \)-intercept at \( 11 \). Connect these points with a smooth downward-opening curve to complete your graph!