Now this? somebody....
1 answer:
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Answer:
y¹⁵-20y¹⁴
Step-by-step explanation:
I'm going to assume that your problem looks like this
which means we have
Answer:
Please find attached sketch of the path of the ball, having plot area and plot points, created with MS Excel
Step-by-step explanation:
Question;
The equation representing the path of the ball obtained from a similar question posted online are;
h₁ = -4·(t + 1)·(t - 5), h₂ = -4·(t - 2)² + 36, h₃ = -4··t² + 16·t + 20
The above equations represent the same path
The equation, h₁ = -4·(t + 1)·(t - 5), gives the roots of the height function, h(t), used in determining the height of the ball after time <em>t</em>
At (t + 1) = 0 (t = -1) or at (t - 5) = 0 (t = 5), the ball is at ground level
The ball reaches the ground, is at ground level at t = 1, and at t = 5 seconds after being tossed, where h(t) = 0
The equation of the path of the ball in vertex form, y = a·(x - 2)² + k, is h₂ = -4·(t - 2)² + 36, where, by comparison, we have;
The vertex of the ball = The maximum height reached by the ball = (h, k) = (2, 36)
The coefficient of the quadratic term, t², is negative, therefore, the shape of the parabola is upside down, ∩, shape
The sketch of the path of the ball created with MS Excel, used in plotting the vertex, the initial value and the root points of the parabola, through which the ball passes and joining of the points with a 'smooth' curve is attached
