Determine the approximate maximum distance
y=-0.01x2 +0.7x+6.1
-100 y = x^2 -70 x - 610
x^2 - 70 x = -100 y + 610
x^2 - 70 x + 1225 = -100y + 1835
(x-35)^2 = -100 y + 1835
vertex at x = 35
then y =18.35
To determine the approximate maximum distance of the given quadratic function y = -0.01x^2 + 0.7x + 6.1, we can use the vertex formula. The vertex formula tells us that the x-coordinate of the maximum point of a quadratic function in the form y = ax^2 + bx + c is given by x = -b / (2a).
In this case, a = -0.01, b = 0.7, and c = 6.1. Plugging these values into the formula, we get:
x = -0.7 / (2 * (-0.01))
= 35
The x-value of the vertex is 35.
To find the corresponding y-value, we substitute this x-value into the original equation:
y = -0.01(35)^2 + 0.7(35) + 6.1
= -0.01(1225) + 24.5 + 6.1
= -12.25 + 24.5 + 6.1
= 18.35
Therefore, the approximate maximum distance is approximately 18.35.