Question

question 4 (multiple choice worth 6 points) (06.03r mc) an archer shoots an arrow up towards a target located on a hill, which is shown by the graph. which set of equations best models the point of intersection of the arrow and the target? $y = -0.008x^{2}+0.2x + 7$ and $y = 0.12x$ $y = 0.008x^{2}+0.2x + 7$ and $y = 0.12x$ $y = -0.008x^{2}+0.2x + 7$ and $y = x$ $y = -0.008x^{2}+0.2x + 7$ and $y = -x$

Explanation:

Step1: Identify arrow's equation shape

The arrow's path is a downward-opening parabola, so the coefficient of $x^2$ must be negative. This eliminates the option with $y = 0.008x^2 + 0.2x + 7$.

Step2: Test intersection point $(35, 4.2)$ on linear equations

First, test $y=0.12x$:
Substitute $x=35$: $y=0.12\times35=4.2$, which matches the point.
Test $y=x$: $y=35
eq4.2$, so this is invalid.
Test $y=-x$: $y=-35
eq4.2$, so this is invalid.

Step3: Verify parabola at intersection point

Substitute $x=35$ into $y=-0.008x^2 + 0.2x +7$:

$$\begin{align*} y&=-0.008\times(35)^2 + 0.2\times35 +7\\ &=-0.008\times1225 +7 +7\\ &=-9.8 +14\\ &=4.2 \end{align*}$$

This matches the $y$-value of the intersection point.

Answer:

A. $y = -0.008x^2 + 0.2x + 7$ and $y = 0.12x$