[tex]a=-1[/tex], because the parabola has the maximum, not the minimum, (∩)
The Vertex form of a parabola equation:
V(-3;4), so:
[tex]f(x)=a(x-p)^2+q[/tex]
[tex]f(x)=-1(x+3)^2+4[/tex]
* [tex]f(x)=-(x+3)^2+4[/tex]
The zeros and the conix form of a parabola equation:
[tex]x_1=-5[/tex]
[tex]x_2=-1[/tex]
[tex]f(x)=a(x-x_1)(x-x_2)[/tex], so:
* [tex]f(x)=-(x+5)(x+1)[/tex]
We can also use the general form of a parabola:
[tex]f(x)=ax^2+bx+c[/tex]
[tex]f(x)=-(x+5)(x+1)=-(x^2+x+5x+5)=-x^2-x-5x-5=-x^2-6x-5[/tex]
* [tex]f(x)=-x^2-6x-5[/tex]
The answers: A, C, E
:)
