Respuesta :
For a quadratic function in the form:
[tex]f(x)=ax^2+bx+c[/tex]If a>0 the function opens up, it has a minimum
If a<0 the function opens down, it has a maximum
Axis of symmetry is x= -b/2a
Vertex: (-b/2a, f(-b/2a)), f(-b/2a) is the maximum or minimum value
For the given function:
[tex]h(t)=-8t^2+2t-1[/tex]a= -8
Parabola opens down. It has a maximum value
Find the axis of symmetry:
[tex]t=-\frac{2}{2(-8)}=\frac{-2}{-16}=\frac{1}{8}[/tex]Find the y-coordinate of the vertex:
[tex]\begin{gathered} h(\frac{1}{8})=-8(\frac{1}{8})\placeholder{⬚}^2+2(\frac{1}{8})-1 \\ \\ h(\frac{1}{8})=-8(\frac{1}{64})+\frac{2}{8}-1 \\ \\ h(\frac{1}{8})=-\frac{1}{8}+\frac{2}{8}-1 \\ \\ h(\frac{1}{8})=\frac{1}{8}-1 \\ \\ h(\frac{1}{8})=\frac{1-8}{8} \\ \\ h(\frac{1}{8})=-\frac{7}{8} \end{gathered}[/tex]______________
Then, the given function has a maximum value, the maximum value is -7/8, and the axis of symmetry is t=1/8