What they are really asking is: Where is the vertex of this parabola?
If you know calculus, it is really easy. You look for where dy/dx = 0
dy/dx = -.5 x + 5
that is 0 when x = 10
then y = -100/4 + 50 or 25
If you do not use calculus then you must put this parabola in standard form by completing the square
-4y = x^2 -20 x
-4y + 100 = x^2 -20 x + 100
-4 (y-25) = (x-10)^2
vertex at (10,25)
as we knew
If the path of a frog's jump is approximated by the equation
y = -0.25x2 + 5x, what is the height of the frog's jump?
3 answers
There are two ways to do this: calculus and completing the square. I don't know if they let you use calculus, so let's do it the other way. "y" is supposed to be the height. First find the value of x for which y is a maximum.
Rewrite as
y = -(x/2)^2 + 5x - (5)^2 + (5)^2 = 0
y = -[(x/2) - 5)]^2 + (5)^2 = 0
The maximum value of y occurs when x = 10, at which time the term in brackets os zero. The value of y them is 25, and that is the maximum.
Rewrite as
y = -(x/2)^2 + 5x - (5)^2 + (5)^2 = 0
y = -[(x/2) - 5)]^2 + (5)^2 = 0
The maximum value of y occurs when x = 10, at which time the term in brackets os zero. The value of y them is 25, and that is the maximum.
Glad I agree with Damon!
1 answer