Question

On a movie set, an archway is modeled by the equation y = -0.5x2 + 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?

Help me!Please!

Answer 1

$y=-0.5x^2+3x \to archway \\ \\ -0.5x + 2.42y = 7.65 \\ \\ y= (7.65+0.5x): 2.42\to y= 0.21x +3.16\to laser \\ \\ \left \{ {y=-0.5x^2+3x \atop { y= 0.21x +3.16}} \right. \\ \\ y=y \\ \\ -0.5x^2+3x= 0.21x +3.16 \\ \\ -0.5x^2+3x- 0.21x -3.16= 0 \\ \\ -0.5x^2+2.79x -3.16= 0 \\ \\ x_ {1\ y\ 2} = \frac{-b\pm \sqrt{b^2-4ac}}{2a}\qquad a=-0.5\quad b=2.79\quad c= -3.16 \\ \\ x_ {1\ y\ 2} = \frac{-2.79\pm \sqrt{(2.79)^2-4(-0.5)(-3.16)}}{2(-0.5)}$

$x_ {1\ y\ 2} = \frac{-2.79\pm \sqrt{7.78-6.32}}{-1} \\ \\ x_ {1\ y\ 2} = \frac{-2.79\pm \sqrt{1.46}}{-1} \\ \\ x_ {1\ y\ 2} = \frac{-2.79\pm 1.21}{-1} \\ \\ x_1= 2.79+1.21\to \boxed{x_1=4} \\ \\ x_2= 2.79-1.21\to \boxed{x_2=1.58} \\ \\ Points \ A \ y\ B \to \\ \\ A=(x, y) \to \boxed{A= (4,4)} \ height = 4\\ \\ B=(x, y) \to \boxed{B= (1.58\ ,3.49)} \ height = 3.49$


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