A frog leaps from a stump 4 feet high and lands 6 feet from teh base of the stump. we can consider the initial position of the frog to be (0,4) and its landing to be (6,0).
the height of the frog in feet is given by the following equation: -1/3x^2+4/3x+4
1. what is the horizontal distance (x) from the base of the stump when the frog reached maximum height?
2. what was the maximum height
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The maximum height is found at the vertex of the parabola. You have a parabola when you have an x^2 term as the highest degree.
The vertex is found at -b/2a this happen to be the horizontal distance. what would -b/2a be?
Are you in Algebra or Calculus? This is an algebraic way of doing it.
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OK.. so let's do it this way.
a = -1/3
b = 4/3
-b/2a = -4/3 over -2/3 ok so far?
ok got ya
When you divide by a fraction, you multiply by the reciprocal.
-4/3 * -3/2 = 12/6 = 2
so x = 2 that is how far away from the stump you are when the the frog is at it's maximum
so the vetex is 2 and thats the horizontal distance?
x = 2
Does my picture make sense... kind of dumb I know
no that really helps
The x value of the vertex is 2 and that is the horizontal distance, now we need to find how high it is at x = 2
Substitute 2 into your equation and find y
or is it y=8
y = 5.3 I think...
(-1/3)*4 + (4/3)*2 +4
-4/3 + 8/3 + 4
4/3 + 4
Notice the vertex (which is the highest point) has coordinates (2,5.3)
ok think i got ya
ok think i got ya
how would u sketch a graph of this problem
Use your xy coordinate graph paper.
put a dot on (0,4) that is the frog.
Put a dot on (6,0) that is where the frog lands
Put a dot on (2,5.3) that is the highest point.
Draw an arc "like throwing a ball up in the air" to the highest point and then down to the (6,0) point. Like a ball would look like if you were throwing it.