Here's my question:
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)
Here's my work on the question:
-x^2+15x=4x-12
x^2-11-12=0
(x-12)(x+1)=0
It needs to be coordinate and can you please show your work for getting the answer,
Thanks you1
To find the point where the fireworks will land on the hill, we need to solve for the x-coordinate from the given equation:
-x^2 + 15x = 4x - 12
Rearranging the equation:
-x^2 + 15x - 4x + 12 = 0
Combining like terms:
-x^2 + 11x + 12 = 0
Now, we will factor this quadratic equation:
(x - 12)(-x - 1) = 0
To find the possible values of x, we set each factor equal to zero:
x - 12 = 0 or -x - 1 = 0
Solving the first equation:
x = 12
Solving the second equation:
-x = 1
x = -1 (multiplying both sides by -1)
So, the possible x-coordinates where the fireworks will land on the hill are x = 12 and x = -1.
To find the corresponding y-coordinate, we substitute these x-values into the equation of the slope of the hill:
y = 4x - 12
For x = 12:
y = 4(12) - 12 = 48 - 12 = 36
For x = -1:
y = 4(-1) - 12 = -4 - 12 = -16
Therefore, the coordinates where the fireworks will land on the hill are (12, 36) and (-1, -16).