Respuesta :
Note: It the given equation the coefficient of [tex]x^2[/tex] must be -16 instead of 16.
Given:
Consider the height of the water balloon over time can be modeled by the function
[tex]y=-16x^2+160x+50[/tex]
To find:
The maximum height of the water balloon after it was thrown.
Solution:
We have,
[tex]y=-16x^2+160x+50[/tex]
Here, leading coefficient is negative. So, it is a downward parabola and vertex of a downward parabola, is the point of maxima.
If a parabola is [tex]f(x)=ax^2+bx+c[/tex], then
[tex]Vertex=\left(-\dfrac{b}{2a},f(-\dfrac{b}{2a})\right)[/tex]
Here, [tex]a=-16,b=160,c=50[/tex]. So,
[tex]-\dfrac{b}{2a}=-\dfrac{160}{2(-16)}[/tex]
[tex]-\dfrac{b}{2a}=-\dfrac{160}{-32}[/tex]
[tex]-\dfrac{b}{2a}=5[/tex]
Now, put x=5 in the given equation.
[tex]y=-16(5)^2+160(5)+50[/tex]
[tex]y=-16(25)+800+50[/tex]
[tex]y=-400+850[/tex]
[tex]y=450[/tex]
The vertex of the given parabolic equation is (5,450).
Therefore, the maximum height of the balloon is 450 units after 5 units of time.
