PLEASE HELP IM GIVING 15 POINTS
On a movie set, an archway is modeled by the equation y = -0.5x^2 + 3x, where y is the height in feet and x is the horizontal distance in feet. A laser is directed at the archway at an angle modeled by the equation -0.5x + 2.42y = 7.65 such that the beam crosses the archway at points A and B. At what height from the ground are the points A and B?

check the picture below.

so, the distance from the ground to points A and B are those two green lines, namely the y-coordinate of their intersection, or solution.

[tex]\bf \begin{cases} \boxed{y}=-0.5x^2+3x\\ -0.5x+2.42y=7.65\\ ----------------\\ -0.5x+2.42\left( \boxed{-0.5x^2+3x} \right)=7.65 \end{cases} \\\\\\ -0.5x-1.21x^2+7.26x=7.65\implies -1.21x^2+6.76x=7.65 \\\\\\ 0=1.21x^2-6.76x+7.65 \\\\\\ ~~~~~~~~~~~~\textit{quadratic formula} \\\\ 0=\stackrel{\stackrel{a}{\downarrow }}{1.21}x^2\stackrel{\stackrel{b}{\downarrow }}{-6.76}x\stackrel{\stackrel{c}{\downarrow }}{+7.65} \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}[/tex]

anyhow, if we run the quadratic formula on those coefficients, we'll end up with x ≈ 4.01023060106514842416, and x ≈ 1.57654625843898380724.

now, what's the y-coordinate for each? well, we can just plug that on say the first equation.

[tex]\bf y=-0.5(4.0102306010651484)^2+3(4.0102306010651484) \\\\\\ y\approx 3.9897170663357744678\\\\ -------------------------------\\\\ y=-0.5(1.5765462584389838)^2+3(1.5765462584389838) \\\\\\ y\approx 3.486889722817971861[/tex]

sqdancefan sqdancefan · Answer 2 · 2018-06-17T22:25:04+02:00

The points of intersection are the points where the (x, y) values satisfy both equations. Since we're interested in the y-values, we can use the second equation to substitute for x in the first equation.

From the second equation,
x - 4.84y = -15.30 . . . . . multiply the second equation by -2
x = 4.84y -15.30 . . . . . . add 4.84y to get an expression for x

Now, we can substitute this expression for x into the first equation.
y = -0.5(4.84y -15.30)² +3(4.84y -15.30)
y = -0.5(23.4256y² -148.104y +234.09) + 14.52y -45.90
Subtracting the right side from the whole equation we get
11.7128y² -87.572y +162.945 = 0

This equation may be best solved using the quadratic formula. For
ax²+bx+c = 0
it tells you the solutions are
x = (-b ±√(b²-4ad))/(2a)

For our equation, with a=11.7128, b=-87.572, c=162.945, the solutions are
y = (87.572 ±√((-87.572)²-4(11.7128)(162.945)))/(2·11.7128)
y = (87.572 ±√34.6864)/23.4256
y ≈ 3.487 ft, 3.990 ft

Points A and B are 3.487 ft and 3.990 ft above the ground.

_____
The numbers might be easier to work with if you solve the second equation for y, then equate the two y expressions. This gives you a quadratic in x. Once you find the solutions for x, you then need to put them back into the equation for y to find the y-values of points A and B. The method used here gives you the answer directly, even if the numbers are a little longer.

Of course, a graphing calculator can help you avoid all the algebra. It will give you the points of intersection without much trouble.