SOLUTION: I am having a really hard time with this problem! Please show me how it should be solved! Thank you in advance! If the path of a frog's jump is approximated by the equation

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I am having a really hard time with this problem! Please show me how it should be solved! Thank you in advance! If the path of a frog's jump is approximated by the equation       Log On


   

Question 162945: I am having a really hard time with this problem! Please show me how it should be solved! Thank you in advance!
If the path of a frog's jump is approximated by the equation y = -0.25x^2 + 5x, what is the height of the frog's jump?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
How high did the frog jump?
y+=+-0.25x%5E2%2B5x
This is a quadratic equation and, when graphed, you will see a parabola that opens downward. The dependent variable (y) represents the height of the frog's jump at time (x) seconds.
You may see this kind of equation written as:
h%28t%29+=+-0.25t%5E2%2B5t
The vertex of the parabola is, of course, a maximum, and this represents the maximum height of the frog's jump.
So you need to find the coordinates of the vertex.
The x-coordinate of the vertex is given by:x+=+-b%2F2a where the a and b come from the general form of the equation:
ax%5E2%2Bbx%2Bc In your problem, a = -0.25 and b = 5, so...
x+=+-5%2F2%28-0.25%29
x+=+10 now substitute this value of x into the original quadratic equation to find the corresponding y-coordinate.
y+=+-0.25%2810%29%5E2%2B5%2810%29 Simplify and solve for y.
y+=+-0.25%28100%29%2B50
y+=+25
The frog's maximum height is 25 feet.
graph%28400%2C400%2C-2%2C24%2C-5%2C30%2C-0.25x%5E2%2B5x%29