−<span>3<span>(<span><span>4a</span>−<span>5b</span></span>)</span></span><span>=<span><span>(<span>−3</span>)</span><span>(<span><span>4a</span>+<span>−<span>5b</span></span></span>)</span></span></span><span>=<span><span><span>(<span>−3</span>)</span><span>(<span>4a</span>)</span></span>+<span><span>(<span>−3</span>)</span><span>(<span>−<span>5b</span></span>)</span></span></span></span><span>=<span><span>−<span>12a</span></span>+<span>15<span>b</span></span></span></span>
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.
Answer:
Step-by-step explanation:
