I’m assuming you want it in slope intercept form so it’s y=1/6x-2
the answer to your question there would be 12
Answer:
x = -8 and x = 4
Step-by-step explanation:
given
f(x) = (x+8) (x - 4)
recall that at any point on the x-axis, y = 0 [i.e f(x) = 0]
hence to find where the graph crosses the x-axis, we simply substitue f(x) = 0 into the equation and solve for x
f(x) = (x+8) (x - 4)
0 = (x+8) (x - 4)
Hence
either,
(x+8) = 0 ----> x = -8 (first crossing point)
or
(x-4) = 0 ------> x = 4 (second crossing point)
Hence the graph crosses the x-axis at x = -8 and x = 4
Problem
For a quadratic equation function that models the height above ground of a projectile, how do you determine the maximum height, y, and time, x , when the projectile reaches the ground
Solution
We know that the x coordinate of a quadratic function is given by:
Vx= -b/2a
And the y coordinate correspond to the maximum value of y.
Then the best options are C and D but the best option is:
D) The maximum height is a y coordinate of the vertex of the quadratic function, which occurs when x = -b/2a
The projectile reaches the ground when the height is zero. The time when this occurs is the x-intercept of the zero of the function that is farthest to the right.
