Fireworks are to be launched from a platform at the base of a hill. Using the top of the platform as the origin and taking some measurements, it was determined that the cross-section of the slope of one side of the hill is y=4x-12 If the path of the fireworks is y=-x^2+15x, calculate the point where the fireworks will land on the hill. (Show your work - 3 marks)
what's the trouble? You have two functions and want to know where they are equal.
-x^2+15x = 4x-12
x^2-11x-12 = 0
(x-12)(x+1) = 0
I think you can take it from there, right?
To find the point where the fireworks will land on the hill, we need to find the intersection point between the path of the fireworks and the cross-section of the slope of the hill.
First, let's set the equations equal to each other to find the intersection point:
-x^2 + 15x = 4x - 12
Rearranging the equation, we get:
-x^2 + 15x - 4x + 12 = 0
Simplifying further, we get:
-x^2 + 11x + 12 = 0
Next, we need to solve this quadratic equation. We can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
For our equation -x^2 + 11x + 12 = 0, the coefficients are:
a = -1, b = 11, c = 12
Plugging these values into the quadratic formula, we get:
x = (-11 ± √(11^2 - 4(-1)(12))) / (2(-1))
Simplifying further, we have:
x = (-11 ± √(121 + 48)) / -2
x = (-11 ± √169) / -2
x = (-11 ± 13) / -2
We have two possible values for x:
x₁ = (-11 + 13) / -2 = -2 / -2 = 1
x₂ = (-11 - 13) / -2 = -24 / -2 = 12
Now that we have the x-values, we can substitute them back into either equation to find the corresponding y-values.
Using the equation y = 4x - 12:
For x = 1, y = 4(1) - 12 = 4 - 12 = -8.
So one point where the fireworks will land on the hill is (1, -8).
For x = 12, y = 4(12) - 12 = 48 - 12 = 36.
So the other point where the fireworks will land on the hill is (12, 36).
Therefore, the fireworks will land on the hill at the points (1, -8) and (12, 36).