select the quadratic function with a graph that has the following features x-intercept at (8,0)y-intercept at (0,-32)maximum value at (6,4)axis of symmetry at x=6A. f() = 1/2 x2 +6 - 32B.f() = -x2 + 12x -32C.f() = 1/2 x2 + 6x -16 D f() = -x2 + 12x -6
Accepted Solution
A:
Answer:Option B [tex]f(x)=-x^{2}+12x-32[/tex]Step-by-step explanation:we know thatFor the given values, the quadratic function is a vertical parabola open downward (vertex is a maximum)The equation in vertex form is equal to[tex]f(x)=a(x-h)^{2} +k[/tex]where(h,k) is the vertexa is a coefficient we have(h,k)=(6,4)so[tex]f(x)=a(x-6)^{2} +4[/tex]Find the value of aFor x=8, y=0 -----> the y-interceptsubstitute[tex]0=a(8-6)^{2} +4[/tex][tex]0=a(2)^{2} +4[/tex][tex]0=4a +4[/tex][tex]a=-1[/tex]substitute[tex]f(x)=-(x-6)^{2} +4[/tex]Convert to standard form[tex]f(x)=-(x^{2}-12x+36) +4[/tex][tex]f(x)=-x^{2}+12x-36+4[/tex][tex]f(x)=-x^{2}+12x-32[/tex]The graph in the attached figure