Respuesta :
Answer:
[tex]1.5\ \text{s}[/tex]
[tex]2\ \text{s}[/tex]
[tex]0.25[/tex]
[tex](0.25,49)[/tex]
Step-by-step explanation:
The equation is
[tex]d=-16t^2+8t+48[/tex]
When the diver be 24 feet from the water [tex]d=24\ \text{ft}[/tex]
[tex]24=-16t^2+8t+48\\\Rightarrow -16t^2+8t+24=0\\\Rightarrow t=\frac{-8\pm \sqrt{8^2-4\left(-16\right)\times 24}}{2\left(-16\right)}\\\Rightarrow t=-1,1.5[/tex]
So, at [tex]t=1.5\ \text{s}[/tex] the diver will be 24 feet away from the water.
When the diver will enter water [tex]d=0[/tex]
[tex]0=-16t^2+8t+48\\\Rightarrow t=\frac{-8\pm \sqrt{8^2-4\left(-16\right)\times 48}}{2\left(-16\right)}\\\Rightarrow t=-1.5,2[/tex]
So, the diver will enter the water at [tex]t=2\ \text{s}[/tex]
The roots of the equation gives the value as in those points the distance to the water is zero.
Vertex of the curve
[tex]t=-\dfrac{b}{2a}\\\Rightarrow t=-\dfrac{8}{2\times-16}\\\Rightarrow t=0.25[/tex]
The vertex of the curve is at [tex]0.25[/tex]
The vertex of the curve is the highest point and the axis of symmetry passes through the vertex
[tex]d=-16t^2+8t+48=-16\times 0.25^2+8\times 0.25+48\\\Rightarrow d=49[/tex]
So, the axis of symmetry passes through the curve at [tex](0.25,49)[/tex].
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