A ball is thrown, and its height is recorded a number of different times. The ball reaches a maximum height of 20 m after 1.5 s.
How do you sketch this relation from just a vertex.
Grade 11 functions
You know the vertex is at (1.5,20)
You know it passes through (0,0)
Use symmetry to draw the other half of the parabola. It also passes through (3,0)
To sketch the relation of the ball's height over time based on the given information, we can use a quadratic function, which has the general form of f(x) = ax^2 + bx + c.
In this case, the vertex form of the quadratic equation is the most appropriate, given that we have the vertex information. The vertex form of a quadratic equation is f(x) = a(x - h)^2 + k, where (h, k) represents the vertex of the parabola.
Given that the ball reaches the maximum height of 20 m after 1.5 seconds, we can conclude that the vertex is located at (1.5, 20). Plugging these values into the vertex form, we obtain the equation f(x) = a(x - 1.5)^2 + 20.
Now, to determine the value of the "a" coefficient, we need another point on the parabola. Since we don't have another specific point, we can use the symmetry property of quadratic functions. According to this property, since the peak is at 1.5 seconds, the ball will also reach the same height after the same amount of time below the peak.
Therefore, if we go 1.5 seconds before the peak, we will reach the same height of 20 m. This means we have another point at (0, 20). Plugging these values into the equation, we can solve for "a."
20 = a(0 - 1.5)^2 + 20
0 = a(2.25)
Since the product of a and 2.25 is zero, we know that the value of a must be zero. Thus, the equation becomes f(x) = 0(x - 1.5)^2 + 20, which simplifies to f(x) = 20.
In conclusion, based on the given information, the relation between the ball's height and time is a horizontal line at a height of 20 m. To sketch this relation, draw a straight line parallel to the time axis (x-axis) that intersects the y-axis at y = 20.