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Brandon Offline
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Difficult Word Problem (College Algebra) - October 7th 2009, 03:26 AM

I have a real difficult time doing word problems because I don't understand how to set the problem up. I've tried for the past 30 minutes, honestly, and I need help desperately!

Here's the problem:

When the shot is released at an angle of 35 degrees, its path can be modeled by the function

f(x) = -0.01x^2 + 0.7x + 6.1

in which x is the shot's horizontal distance, in feet, and f(x) is its height, in feet. Use the function do determine the shot's maximum distance.

The only thing I've came up with is that since x represents the horizontal distance, I need to get x by itself to figure out what the horizontal distance is. Other than that, I am absolutely 100% stumped. If you understand the problem, don't tell me what the answer is...all I'm concerned about is how you would work out the problem.

The next problem I have is just solving for x. (1/x) + (1/(x+2)) = (1/3). The denominators are x, x+2, and 3...I know that, but there's no common denominator I don't think...so I also don't know how to solve this problem.

I'd appreciate your help.
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Re: Difficult Word Problem (College Algebra) - October 7th 2009, 04:24 AM

It's a parabola, right? Well, the furthest it can go is when is hits the x axis, right? Plug zero in for f(x). Whatever "X" is, is your answer... I'm pretty sure.

You can check it on a graphing calculator. I'm like 99% the x-intercep is the answer.
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Re: Difficult Word Problem (College Algebra) - October 7th 2009, 07:28 AM

1/x) + (1/(x+2)) = (1/3)

You need to multiply the numerator and denominator of each fraction by the denominators of the other two fractions. So you get:

[3(x+2)]/[3(x)(x+2)] + [3(x)]/[3(x)(x+2)] = [(x+2)(x)]/[3(x)(x+2)]

Then you want to expand the numerators:

(3x + 6)/[3(x)(x+2)] + 3x/[3(x)(x+2)] = [(x^2) + 2x]/[3(x)(x+2)]

Now you want to move everything to one side, setting the equation equal to 0:

[3x + 6 + 3x - x^2 - 2x]/[3(x)(x+2)] = 0

Combine like terms, and you'll get:

[-x^2 + 4x + 6]/[3(x)(x+2)] = 0

The fraction equals 0 when the numerator equals 0, so what you need to solve is:

[-x^2 + 4x + 6] = 0

To find the roots of this equation, you'll need to use the quadratic equation,

x = [-b + √(b^2 - 4ac)]/2a and x = [-b - √(b^2 - 4ac)]/2a

x = [-4 + √(-4^2 - 4(-1)(6))]/2(-1) and x = [-4 - √(-4^2 - 4(-1)(6))]/2(-1)

x = [-4 + √(16+24)]/-2 and x = [-4 - √(16+24)]/-2

x = [-4 + √40]/-2 and x = [-4 - √40]/-2

x = [-4 + 2√10]/-2 and x = [-4 - 2√10]/-2

For your first problem, you need to find the roots of the equation. If we're thinking about this problem, the model only makes sense when the object has a height of greater than or equal to zero. This occurs between the two roots of the equation. It also only makes sense when distance is greater than 0, because you cannot have a negative distance. You can use the quadratic formula from the above problem, and you should get one positive answer, and one negative. A negative answer doesn't make sense for this problem, because you can't have a negative distance. So the maximum height is the positive root.

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Last edited by Grizabella; October 7th 2009 at 06:11 PM.
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Re: Difficult Word Problem (College Algebra) - October 7th 2009, 04:31 PM

Jessica did a good a good job explaining the second one, but the first one I didn't get it.

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Re: Difficult Word Problem (College Algebra) - October 10th 2009, 09:09 PM

For the first problem, just picture what the parabola will look like in your head. Since the first term is negative, it will open downward and the parabola will look like the path of a projectile. The maximum distance is simply just the rightmost root of the parabola. So just use the quadratic formula (since the coefficients are messy) to find the two roots and just choose the bigger of the two roots. That will be the maximum distance.
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