Respuesta :
To answer this question, we can see that the function that gives the height of the weight at any given time is a parabola with a maximum (the negative number in front of the leading term gives us this information):
[tex]h(t)=-16t^2+32t+3[/tex]We can find the maximum height of the weight by finding the vertex of the parabola - we already know that this vertex is a maximum in this case.
The formula to find the vertex of a parabola is given by:
[tex]\begin{gathered} x_v=-\frac{b}{2a} \\ y_v=c-\frac{b^2}{4a} \end{gathered}[/tex]This is for the next general equation:
[tex]ax^2+bx+c[/tex]Now, we have that the parabola is:
[tex]h(t)=-16t^2+32t+3[/tex]Then we have:
[tex]\begin{gathered} a=-16 \\ b=32 \\ c=3 \end{gathered}[/tex][tex]\begin{gathered} x_v=-\frac{b}{2a} \\ x_v=-\frac{32}{2(-16)}=-\frac{32}{-32}=1 \\ x_v=1 \end{gathered}[/tex]And
[tex]\begin{gathered} y_v=c-\frac{b^2}{4a} \\ y_v=3-\frac{(32)^2}{4(-16)} \\ y_v=3-\frac{32^2}{-64} \\ y_v=3+\frac{32\cdot32}{2\cdot32}=3+\frac{32}{2}\cdot\frac{32}{32}\Rightarrow\frac{32}{32}=1,\frac{a}{a}=1 \\ y_v=3+\frac{32}{2} \\ y_v=3+16=19 \\ y_v=19 \end{gathered}[/tex]Therefore, we have that the vertex of the parabola is (1, 19). That is, we have that the maximum point for this parabola is y = 19 - this is the maximum value for the function, and the function takes this value when x = 1.
In other words, in summary, we have:
• The maximum height of the weight is 19 feet (we assume the function gives us the height in feet.)
,• The weight will take 1 second to reach the maximum height (we assume that the function is expressed in seconds.)
,• The contestant will not win a prize since the weight will never reach 20 feet.
