Question

1. the deadly woodland hopping mouse has terrorized the people of west virginia for close to a decade. it is believed that a fanatic university of virginia football fan and noted ecologist introduced the mouse to the forest of west virginia after the west virginia football team defeated the virginia team in their annual game. since then, the mouse has wiped out the entire cattle trade and is responsible for the deaths of 284 people. many more deaths are suspected to have been caused by the mouse but have not been officially laid at his feet because there were no survivors to recount the carnage. the vicious killer gets its name from its uncanny ability to leap great distances when attacking its prey. in an eerie similarity to the movie cloverfield, the government has documentary footage from one mouse attack on a family vacationing in a west virginia state park. military mathematicians have analyzed the footage and have estimated that the flight path of the mouses assault on the fleeing oldest son in the video can be modeled by the equation:$y = -0.2x^{2}+4x - 12$graph the equation in desmos. the x-axis represents the ground that the mouse and boy are running along. even though the parabola falls below the x-axis, we only want to view the part that models the mouses jump. the y-axis represents the height of the mouse off the ground.- does the mouse jump from (0,0)? what could be an explanation for this?- how far in feet can this deadly killer leap?- the son in the video is 6ft tall. how far did the mouse jump when it attacked him by landing on top of his head?- what is the maximum height the mouse reaches in its leap?

Explanation:

Step1: Check if (0,0) is on the curve

Substitute $x=0$ into $y=-0.2x^2+4x-12$:
$y=-0.2(0)^2+4(0)-12=-12$

Step2: Explain non-zero y at x=0

Since $y=-12$ when $x=0$, the point $(0,0)$ is not on the parabola. This means the mouse does not jump from ground level (x-axis) at the start; its starting position is 12 feet below the defined ground reference, so it must jump from an elevated position like a tree or rock to match the model.

Step3: Find horizontal leap distance

The leap ends when the mouse lands on the ground ($y=0$). Solve $-0.2x^2+4x-12=0$:
Multiply through by -5: $x^2-20x+60=0$
Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=1, b=-20, c=60$:
$x=\frac{20\pm\sqrt{400-240}}{2}=\frac{20\pm\sqrt{160}}{2}=\frac{20\pm4\sqrt{10}}{2}=10\pm2\sqrt{10}$
Calculate the two roots: $x_1=10+2\sqrt{10}\approx16.32$, $x_2=10-2\sqrt{10}\approx3.68$
Leap distance = $x_1-x_2=(10+2\sqrt{10})-(10-2\sqrt{10})=4\sqrt{10}\approx12.65$ feet

Step4: Find jump distance at y=6

Set $y=6$ and solve for $x$:
$6=-0.2x^2+4x-12$
Rearrange: $0.2x^2-4x+18=0$
Multiply by 5: $x^2-20x+90=0$
Quadratic formula: $x=\frac{20\pm\sqrt{400-360}}{2}=\frac{20\pm\sqrt{40}}{2}=10\pm\sqrt{10}$
Distance = $(10+\sqrt{10})-(10-\sqrt{10})=2\sqrt{10}\approx6.32$ feet

Step5: Find maximum height of parabola

For $y=ax^2+bx+c$, vertex x-coordinate is $x=-\frac{b}{2a}$:
$x=-\frac{4}{2(-0.2)}=10$
Substitute $x=10$ into the equation:
$y=-0.2(10)^2+4(10)-12=-20+40-12=8$

Answer:

1. The mouse does NOT jump from (0,0). When $x=0$, $y=-12$, meaning the model starts 12 feet below the defined ground reference, so the mouse must jump from an elevated position (like a tree) to fit the parabola.
2. The mouse can leap $4\sqrt{10}\approx12.65$ feet horizontally.
3. When landing on the 6 ft tall boy, the mouse jumps $2\sqrt{10}\approx6.32$ feet horizontally.
4. The maximum height the mouse reaches is 8 feet.