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marysya [2.9K]
3 years ago
11

A football is kicked toward the goal. The height of the ball is modeled by the function h(t) = −16t2 + 64t where t equals the ti

me in seconds and h(t) represents the height of the ball at time t seconds. What is the axis of symmetry, and what does it represent?

Mathematics
1 answer:
kondor19780726 [428]3 years ago
8 0

check the picture below.


so, since the ball kick is a parabola opening downwards, like you see in the picture, the axis of symmetry will come from the x-coordinate of the vertex, so let's find it.


\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{64}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{\textit{0, since it was on the ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-16t^2+64t\implies h(t)=-16t^2+64t+0 \\\\[-0.35em] \rule{34em}{0.25pt}


\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+64}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{64}{2(-16)}~,\qquad \qquad \right)\implies (2~,~\qquad )~\hfill \stackrel{\textit{axis of symmetry}}{x=2} \\\\\\ ~\hspace{34em}

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Answer:

177.3 feet

This is a classic find the vertex of a parabola question.

if this was a calculus class the solution would be to take the derivative and set it equal to zero... -32t+ 105 = 0

BUT i assume that you are not in a calculus class..

so we try plan "B" the  highest (or lowest) point of parabola is it's vertex

the vertex formula is [-b/2a,f(-b/2a)]

in your problem a = -16, b=105, c= 5

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<h2>Answer</h2>

2. The Graph Contains The Point (5, 12)

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Remember that in a linear function of the form y=mx+b, m is the slope/rate of change of the line and b is the y-intercept.

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