Answer :
Given:
[tex]y=-2x^2+4x+11[/tex]Required:
(a) Find parabola opens upward.
(b) Find the vertex.
(c) Find the X-intercepts.
(d) Find Y intercepts.
Explanation:
The given equation is:
[tex]y=-2x^{2}+4x+11[/tex]Rewrite it as:
[tex]\begin{gathered} y=-2(x^2-2x)+11 \\ y=-2(x^2-2x+1)+11+2 \\ y=-2(x-1)^2+13 \end{gathered}[/tex]Compare the equation with the standard equation
[tex]y=a(x-h)^2+k[/tex]a = -2, h =1 and k =13
(a) Since the value of a = -2 < 0 so the parabola opens down.
(b) The vertex of the parabola is (h,k) .
Thus the vertex of given the parabola is (1, 13).
(c) At the X-intercept y = 0
[tex]\begin{gathered} -2x^2+4x+11=0 \\ 2x^2-4x-11=0 \end{gathered}[/tex]The given equation is a quadratic equation.
Compare it with
[tex]ax^2+bx+c=0[/tex]solve it as
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex](d) At the y-intercept x =0
[tex]\begin{gathered} y=-2(0)^2+4(0)+11 \\ y=11 \end{gathered}[/tex]Final Answer:
(a) Parabola opens downward
(b) vertex = (1, 13)
(d) y-intercept is 11