Landon is standing in a hole that is 5.1 ft deep. he throws a rock, and it goes up into the air, out of the hole, and then lands on the ground above. the path of the rock can be modeled by the equation y = -0.005x^2 + 0.41x - 5.1, where x is the horizontal distance horizontal distance of the rock, in feet, from landon and y is the height, in f of the rock, in feet, from landon and y is the height, in feet, of the rock above the ground. how far horizontally from landon will the rock land? 33.35 ft 66.71 ft 7.65 ft 15.29 ft

When the rock lands on the ground, it has no height. Or you could say that it has a height of 0. Therefore, our equation when the height, y, is 0, looks like this: . Where the height is 0 is where the quadratic goes through the x axis. To find those places where the graph goes through the x-axis, or where the rock hits the ground, x, we have to factor that quadratic to solve for x. Fit that into the quadratic formula to find that x = 15.29 and x = 66.71. These 2 points are the x-intercepts. Because Landon is below the ground, which also happens to be below the first x-intercept, this means that when he throws the rock it goes UP through x = 15.29, up to its highest point (the vertex), and then lands at the other x-intercept, which is 66.71. The x values are where the ROCK is at ground level, not Landon. So your answer is 66.71.

2.83941151 would be the answer you're looking for! : )

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.