SOLUTION: HELP with this graphing problem please https://imgur.com/a/ndjmykk

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

You can use a graphing tool such as GeoGebra or Desmos to graph y+=+40-5x-5x%5E2 (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
x+=+%28-b%2Bsqrt%28b%5E2-4ac%29%29%2F%282a%29 or x+=+%28-b-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-%28-5%29%2Bsqrt%28%28-5%29%5E2-4%28-5%29%2840%29%29%29%2F%282%28-5%29%29 or x+=+%28-%28-5%29-sqrt%28%28-5%29%5E2-4%28-5%29%2840%29%29%29%2F%282%28-5%29%29

x+=+%285%2Bsqrt%28825%29%29%2F%28-10%29 or x+=+%285-sqrt%28825%29%29%2F%28-10%29

x+=+%285%2B28.7228132326901%29%2F%28-10%29 or x+=+%285-28.7228132326901%29%2F%28-10%29

x+=+%2833.7228132326901%29%2F%28-10%29 or x+=+%28-23.7228132326901%29%2F%28-10%29

x+=+-3.37228132326901 or x+=+2.37228132326901

x+=+-3.372 or x+=+2.372

We ignore the negative x value because we cannot have a negative time value. The only practical solution is roughly x+=+2.372 which rounds to 2.37 as we found earlier.

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Answer: 2.37

Answer by greenestamps(13203) About Me  (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.