Keisha is an Olympic diver. One of Keisha’s dives can be modelled by the relation y=-5x^2+6x+3, where y represents keishas height above the water, in metres, and x represents time in seconds. From what height does Keisha dive? What was keishas maximum height? When did Keisha reach her maximum height? How long did it take Keisha to reach the water.
To find the initial height from which Keisha dives, we can look at the constant term in the equation given. The constant term is 3, so Keisha dives from a height of 3 meters above the water.
To find Keisha's maximum height, we need to first convert the given equation into vertex form. The vertex form of a quadratic equation is y = a(x-h)^2 + k, where (h,k) is the vertex of the parabola.
In this case, the equation y = -5x^2 + 6x + 3 can be rearranged to y = -5(x^2 - (6/5)x) + 3. Completing the square inside the parentheses, we get y = -5(x^2 - (6/5)x + 9/25 - 9/25) + 3. Simplifying further, we get y = -5(x-3/5)^2 + 27/5.
From this form, we can see that the vertex of the parabola is at (3/5, 27/5). This means that Keisha's maximum height is 27/5 meters.
To find when Keisha reaches her maximum height, we can use the x-coordinate of the vertex. Keisha reaches her maximum height at x = 3/5 seconds.
Finally, to find out when Keisha reaches the water, we need to find the x-intercepts of the equation. Setting y to 0, we get 0 = -5x^2 + 6x + 3. Using the quadratic formula, we get x = (-6 ± √(6^2 - 4*(-5)*3))/(2*(-5)). Calculating this, we get x = -0.6 and x = 1. Calculating the positive value, we get x = 1 second.
Therefore, Keisha takes 1 second to reach the water.