Answer:
1) [tex]x=\dfrac{5}{2}[/tex]
2) [tex]t_1=-1[/tex] and [tex]t_2=6[/tex]
3) (0,12)
4) [tex]\left(\dfrac{5}{2},\dfrac{49}{2}\right)[/tex]
Step-by-step explanation:
The function [tex]h(t) =-2t^2 + 10t + 12[/tex] represents the height (h) of the ball after t seconds.
Find the vertex of the parabola:
[tex]t_v=-\dfrac{b}{2a}=-\dfrac{10}{2\cdot (-2)}=\dfrac{5}{2}\\ \\h_v=h\left(\dfrac{5}{2}\right)=-2\cdot\left(\dfrac{5}{2}\right)^2+10\cdot \dfrac{5}{2}+12=-2\cdot \dfrac{25}{4}+25+12=-\dfrac{25}{2}+37=\dfrac{49}{2}[/tex]
Hence, the vertex of the function is at point [tex]\left(\dfrac{5}{2},\dfrac{49}{2}\right).[/tex]
The axis of symmetry of the function is vertical line which passes through the vertex, so its equation is
[tex]x=\dfrac{5}{2}[/tex]
To find y-intercept, equat t to 0 and find h:
[tex]h=-2\cdot 0^2+10\cdot 0+12=12[/tex]
Hence, y-intercept is at point (0,12)
To find the roots of the quadratic finction, equate h to 0 and solve the equation for t:
[tex]h=0\Rightarrow -2t^2+10t+12=0\\ \\t^2-5t-6=0\ [\text{Divided by -2}]\\ \\D=(-5)^2-4\cdot 1\cdot (-6)=25+24=49\\ \\t_{1,2}=\dfrac{-(-5)\pm \sqrt{49}}{2\cdot 1}=\dfrac{5\pm 7}{2}=6,\ -1[/tex]
Therefore, two roots are [tex]t_1=-1[/tex] and [tex]t_2=6[/tex]
