13. you launch an angry bird through the air. the function f(t)= - 10t² + 100t represents the height of the…

Explanation:

Step1: Identify the function form

The function $f(t)= - 10t^{2}+100t$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-10$, $b = 100$, $c = 0$.

Step2: Find the time of maximum height

The $t$-coordinate of the vertex of a quadratic function $y=ax^{2}+bx + c$ is given by $t=-\frac{b}{2a}$. Substituting $a=-10$ and $b = 100$ into the formula, we have $t=-\frac{100}{2\times(-10)}=\frac{-100}{-20}=5$ seconds.

Step3: Find the maximum - height

Substitute $t = 5$ into the function $f(t)=-10t^{2}+100t$. Then $f(5)=-10\times5^{2}+100\times5=-10\times25 + 500=-250+500 = 250$ inches.

Step4: Check if the bird hits the target

First, convert 1 foot to inches. Since 1 foot = 12 inches. We need to find if there is a solution for $f(t)=12$, so set $-10t^{2}+100t=12$. Rearrange it to the standard quadratic - form $10t^{2}-100t + 12=0$ or $5t^{2}-50t+6 = 0$. The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 5$, $b=-50$, $c = 6$. Calculate the discriminant $\Delta=b^{2}-4ac=(-50)^{2}-4\times5\times6=2500 - 120=2380$. Then $t=\frac{50\pm\sqrt{2380}}{10}=\frac{50\pm2\sqrt{595}}{10}=5\pm\frac{\sqrt{595}}{5}\approx5\pm4.88$. The solutions are $t_1\approx5 + 4.88=9.88$ and $t_2\approx5 - 4.88 = 0.12$. So the bird will hit the target.

Step5: Graph the function

For the $x$-intercepts, set $y = f(t)=0$. So $-10t^{2}+100t=0$. Factor out $-10t$: $-10t(t - 10)=0$. The $x$-intercepts are $t = 0$ and $t = 10$. The vertex is at $(5,250)$. Scale the $t$-axis (time) in seconds and the $f(t)$-axis (height) in inches.

Answer:

a) 5 seconds b) Graph with $x$-intercepts at $t = 0$ and $t = 10$, vertex at $(5,250)$. Scale the $t$-axis in seconds and the height - axis in inches. c) 250 inches d) Yes, since the quadratic equation $5t^{2}-50t + 6=0$ has real solutions $t\approx0.12$ and $t\approx9.88$.