SOLUTION: HELP with this graphing problem please
https://imgur.com/a/ndjmykk
Algebra.Com
Question 1182154: HELP with this graphing problem please
https://imgur.com/a/ndjmykk
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
You can use a graphing tool such as GeoGebra or Desmos to graph (note I'm using x in place of t)
Then click on the location where the curve crosses the x axis to determine the coordinates of that point. The location is approximately (2.372, 0)
You'll only worry about cases when x > 0.
Desmos Link
https://www.desmos.com/calculator/hlyqda56a1
That's the graph I set up. It's basically a parabola that's upside down.
The x coordinate of the x intercept is what we want. The 2.372 rounds to 2.37 which is the only answer.
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If you're not allowed to use a graphing tool, then the quadratic formula is the next best thing
y = 40 - 5x - 5x^2
y = -5x^2 - 5x + 40
We have something in the form y = ax^2+bx+c where
a = -5
b = -5
c = 40
Those three values are plugged into the quadratic formula below
or
or
or
or
or
or
or
We ignore the negative x value because we cannot have a negative time value. The only practical solution is roughly which rounds to 2.37 as we found earlier.
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Answer: 2.37
Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!
When the object hits the ground, its height is 0.
Set h=0 in the given equation and solve for t.
It doesn't look as if the quadratic is going to factor, so use the quadratic formula or a graphing calculator.
Obviously there will only be one time when the ball hits the ground. The quadratic equation will have a negative root; but obviously that root makes no sense in the given problem.
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