SOLUTION: find the vertex using completing the square on the equation y=-4.9t^2+51t+1.3 you are trying to find the maximum height

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Question 1205580: find the vertex using completing the square on the equation
y=-4.9t^2+51t+1.3
you are trying to find the maximum height
Found 5 solutions by MathLover1, MathTherapy, greenestamps, ikleyn, math_tutor2020:
Answer by MathLover1(20849) About Me  (Show Source):
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y=-4.9t%5E2%2B51t%2B1.3
y=-4.9%28t%5E2-%2851%2F4.9%29t%29%2B1.3
y=-4.9%28t%5E2-10.408t%2Bb%5E2%29-%28-4.9b%5E2%29%2B1.3...............b=10.408%2F2=5.204
y=-4.9%28t%5E2-10.408t%2B5.204%5E2%29%2B4.9%2A5.204%5E2%2B1.3
y=-4.9%28t-5.204%29%5E2%2B132.6999184%2B1.3
y=-4.9%28t-5.204%29%5E2%2B133.9999184

h=5.204
k=133.9999184
k=134 (approximately)
vertex is at: (5.204,+134)
the maximum height: 134


Answer by MathTherapy(10552) About Me  (Show Source):
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find the vertex using completing the square on the equation
y=-4.9t^2+51t+1.3

you are trying to find the maximum height

It makes absolutely no sense completing the square to determine the vertex. 

All you need to do is apply the formula for the x-coordinate of the vertex/axis of symmetry (matrix%281%2C3%2C+t%2C+%22=%22%2C+%28-+b%29%2F%282a%29%29). 
Then, substitute that "t" value into the given quadratic equation to get the y-coordinate of the vertex,
which happens to be the MAXIMUM height. This you also need.

Then again, you may be learning how to complete the square and this is an exercise to familiarize yourself with
it and go through the process.

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


For a quadratic equation like this, with "ugly" coefficients, the best way to find the maximum height is with a graphing calculator or some other graphing tool.

I definitely agree with tutor @MathTherapy that using completing the square to find the maximum height with this equation is absurd.

In theory, completing the square is a useful skill. But practice for the student in the process of completing the square should be with equations that have "nice" coefficients.

Even the task of finding the maximum height using the fact that the x-coordinate of the vertex is -b/2a and plugging that value into the equation to find the y-coordinate of the vertex is extremely awkward with an equation like this one.

Give the student practice finding the vertex by completing the square or by using x=-b/2a using "nice" equations; for an equation like this, use a graphing calculator.

The instruction to work this problem by completing the square is absurd.

Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.

In this site,  there is a bunch of lessons on a projectile thrown/shot/launched vertically up
with the task to find the maximum height

    - Problem on a projectile moving vertically up and down
    - Problem on an arrow shot vertically upward
    - Problem on a ball thrown vertically up from the top of a tower
    - Problem on a toy rocket launched vertically up from a tall platform
    - A flare is launched from a life raft vertically up

Consider these lessons as your textbook,  handbook,  tutorials and  (free of charge)  home teacher.
Read them attentively and learn on how to solve this type of problems once and for all.



Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

I'll use x in place of t.

The equation
y=-4.9t^2+51t+1.3
becomes
y=-4.9x^2+51x+1.3

You have a few options when it comes to determining the value of x that leads to the max height. This special x value is the axis of symmetry.
  • Use the formula x = -b/(2a) to determine the axis of symmetry (recommended)
  • Completing the square. Perhaps not recommended since it takes a while and could be error-prone.
  • Use the quadratic formula to find the two roots and then apply the midpoint. As the name "axis of symmetry" implies, it's at the midpoint of the roots. This method may be error-prone so it's probably not recommended compared to the 1st option.
  • Use graphing technology such as: GeoGebra, Desmos, TI84, etc. Make sure you use x instead of t. Desmos is perhaps the most easiest to work with since you can click on the max point to display its coordinates.
After determining the value of x (or t), plug it into the given equation to find y.
You should find that the max occurs at the approximate location (5.2040817063, 134.0040816327)

When rounding to say 3 decimal places, that would be (5.204, 134.004)

Therefore, the max height is roughly 134.004 feet.