The function y=-0.03(x-14)^2+6 models the jump of a red kangaroo where x is the horziontal distance (in feet) and y is the corresponding height (in feet).
a)What is the kangaroo's maximum height (in feet)?
b)How long is the kangaroo's jump?
The function y=-0.03(x-14)^2+6 models the jump of a red kangaroo where x is the horziontal distance (in feet) and y is the corresponding height (in feet).
a)What is the kangaroo's maximum height (in feet)?
b)How long is the kangaroo's jump?
Hi Jackie,
I've edited part b since I realized after looking at other entries that we're looking at the path traced by a single jump. Part b is also a little strange since the kangaroo jumps from a negative starting distance; I've chosen to interpret that as part of the jump.
a) 6 b) about 28.28
For Algebra II I think you would look at this problem as a series of transformations of the graph x^2. A regular graph of x^2 looks like a U with the bottom of the U at (0,0). The graph of (x-14)^2 looks like a U moved over to the right, so the bottom of the U is at (14,0). The graph of -.03(x-14)^2 flips the U upside down and stretches it so the TOP of the upside-down U is now at (14,0). Finally the 6 moves the whole graph up by 6 units, so now the top of the upside-down U is at (14,6). So if you're looking at an upside down U, the highest point is (14,6). So 6, the highest y coordinate, is the answer to part a).
For part b), we have to look at the where the function crosses the x axis. So we set your function equal to 0 and solve for two values of x, then take their difference. We have -.03x(x-14)^2 +6=0, then -.03(x-14)^2=-6, then (x-14)^2=200, then x-14=200^.5, then x=(200^.5)+14. So x can be about -.14 or 28.14. I chose to find the difference of these two values, so for part b) the kangaroo jumps about 28.28 feet. **Note: If the kangaroo really started from x= -.14 feet, as in the case where someone had set up a starting line and the kangaroo jumped a little before that line (what an unruly kangaroo!), it would have made more sense to later shift the graph to start at x=0.
Sincerely, Sonja
Hopefully, Sung taee's answer will show up at some point (I have experienced the same problem in the past). In case it doesn't, I will answer this as well.
In order to solve this problem, I am going to work it backwards. That is, I am going to solve part b, then solve part a.
In order to find how long the the kangaroo's jump was, we are going to find out what the value of x is when y=0, because y=0 represents the kangaroo on the ground.
0 = -0.03(x - 14)^{2} + 6 Original equation with 0 substituted for y
0.03(x - 14)^{2} = 6 Add 0.03(x - 14)^{2} to each side
(x - 14)^{2} = 200 Divide both sides by 0.03
x - 14 = +/- 10√2 Take the square root of each side
x = 14 +/- 10√2 Add 14 5o each side
Because 10√2 is approximately 14.14, 14 - 10√2 is approximately 0. This makes sense because the kangaroo has not left the ground yet at 0 ft. The correct answer is 14 + 10√2, which is approximately 28.14 ft.
For part a, we want to use the distance that is halfway between the zeroes of the jump. That is the point halfway between 14 - 10√2 and 14 + 10√2. In other words, we need to use the value 10√2 (approx. 14.14) for x and solve for y.
y = -0.03(10√2 - 14)^{2} + 6 Write original equation with 10√2 substituted for y
y = -0.03(14.14 - 14)^{2} + 6 (Using approximation)
y = -0.03(0.14)^{2} + 6 Substitute (14.14 - 14 = 0.14)
y = -0.03(0.0196) + 6 Square 0.14
y = -0.000588 + 6 Multiply -0.03 times 0.0196
y = 5.999412 Add -0.00588 and 6
y = 6 Round
I went through all of the this work to show that the maximum point of the parabola (the highest point in the kangaroo's jump) is actually the point where the value being squared is 0 (14 - 14). The y-value of the vertex is 6. That is also the highest point in the jump.
The answers are:
a. 6 ft
b. 28.14 ft
Dear John
Let me explain about the given situations.
for b), I think you have to think about the characteristics of parabolar equation.
This parabolar equation has zeros at 14-10v2 and 14+10v2 , therefore, the center or vortex of this equation is located on x = ( (14-10v2 )+( 14+10v2) )/ 2 = 14.00 . Therefore, even though the answer is the same, the approaching method has a problem. ** as I told, the kangaroo did not jump at (0,0) point it jumped at (0, 0.12) point.
I hope you have a good day.
I agree with your analysis Sung taee. I kind of switched between solving the problem in terms of Physics definitions and in terms of the actual equation given halfway through. I definitely agree with your comment.
Hi Jackie
a) If you differentiate the given equation, y' = -0.06(x-14). The maximum point(absolute maximum in this problem) will be when y' = 0. y'=0 = -0.06(x-14). Therefore, x = 14 when y has maximum which is 6.
So, the red kangaroo will jump upto 6 feet when horizontal distance is 14 feet.
b) If y represent height from the surface of ground, the x axis represent the groound and it must be flat.
Then, when the kangaroo will be come back to ground after jump when y = 0.
Therefore, y = 0 = -0.03(x-14)^2 + 6.
-6 = -0.03(x-14)^2
200 = (x-14)^2
10√2 = x-14
x = 14+10√2 = 28.14 [feet] . So the red kangaroo can hop upto 28.14 feet .
** Graphically, this equation has concave down condition. Therefore, it will have absolutely maximum value at the vortex point which is (14, 6). ** At x = 0, the y = 0.12. This means that the kangaroo is not jumped on a horizontal line at time t=0. It jumped at 0.12 feet height at the origine. This is why the distance it jumped is not 28 feet.
Hope this is helpfull.
Sung, while your method works fine, this is an algebra II problem, which means the student probably has not done differentiation yet.
It would be better to solve (a) by looking at the equation as it's in vertex form:
y= a(x-h)2+k, where the vertex is at (h, k)
y=-0.03(x-14)^2+6 so the vertex is (14, 6), so the max. height is 6ft.
That should be y= a(x-h)^{2}+k for the vertex form.
Dear Daniel O.
Thank you for your comment.
Sung
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