Answer:
The vertex is the point (-1,11). is a maximum
Step-by-step explanation:
we know that
The equation of a vertical parabola in vertex form is equal to
[tex]f(x)=a(x-h)^{2}+k[/tex]
where
(h,k) is the vertex
a is a coefficient
if a > 0 the parabola open upward and the vertex is a minimum
if a < 0 the parabola open downward and the vertex is a maximum
we have
[tex]f(x)=-5x^{2}-10x+6[/tex]
Convert to vertex form
Complete the square
[tex]f(x)-6=-5x^{2}-10x[/tex]
Factor the leading coefficient
[tex]f(x)-6=-5(x^{2}+2x)[/tex]
[tex]f(x)-6-5=-5(x^{2}+2x+1)[/tex]
[tex]f(x)-11=-5(x^{2}+2x+1)[/tex]
Rewrite as perfect squares
[tex]f(x)-11=-5(x+1)^{2}[/tex]
[tex]f(x)=-5(x+1)^{2}+11[/tex]
The vertex is the point (-1,11)
The coefficient a=-5
so
a < 0 the parabola open downward and the vertex is a maximum
