Question

Question1: A ball is shot from a cannon into the air with an upward velocity of 10 ft/sec. The function h(t) =-2t^2 + 10t + 12= represents the height (h) of the ball after t seconds. 1. Determine what is the axis of symmetry of the function? 2. Identify the roots of the quadratic function? 3. What is the y-intercept of the quadratic function? .4.) Determine the vertex of the function?

Answer 1

Answer:

1) $x=\dfrac{5}{2}$

2)  $t_1=-1$ and $t_2=6$

3) (0,12)

4) $\left(\dfrac{5}{2},\dfrac{49}{2}\right)$

Step-by-step explanation:

The function $h(t) =-2t^2 + 10t + 12$ represents the height (h) of the ball after t seconds.

Find the vertex of the parabola:

$t_v=-\dfrac{b}{2a}=-\dfrac{10}{2\cdot (-2)}=\dfrac{5}{2}\\ \\h_v=h\left(\dfrac{5}{2}\right)=-2\cdot\left(\dfrac{5}{2}\right)^2+10\cdot \dfrac{5}{2}+12=-2\cdot \dfrac{25}{4}+25+12=-\dfrac{25}{2}+37=\dfrac{49}{2}$

Hence, the vertex of the function is at point $\left(\dfrac{5}{2},\dfrac{49}{2}\right).$

The axis of symmetry of the function is vertical line which passes through the vertex, so its equation is

$x=\dfrac{5}{2}$

To find y-intercept, equat t to 0 and find h:

$h=-2\cdot 0^2+10\cdot 0+12=12$

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:

$h=0\Rightarrow -2t^2+10t+12=0\\ \\t^2-5t-6=0\ [\text{Divided by -2}]\\ \\D=(-5)^2-4\cdot 1\cdot (-6)=25+24=49\\ \\t_{1,2}=\dfrac{-(-5)\pm \sqrt{49}}{2\cdot 1}=\dfrac{5\pm 7}{2}=6,\ -1$

Therefore, two roots are $t_1=-1$ and $t_2=6$