Solving Curve Function: Intercepts, Parabola Help

[cgaday]

I am trying to figure out this curve function for the given data.

y intercept = 1
i have x intercept values of 4 and -4,

it is a parablola.

I can remember how to integrate, but when drew out the curve I could not figure out the function of it for the life of me.

Let me know if you need more info. thanks.

 

[Gib Z]

You know the equation is a quadratic, and you know its two roots. Not too hard if you think about it.

 

[cgaday]

so x=-4 and x=4.

then i get (x+4)(X-4) = x^2 -16

then what with the y intercept.

it's been a long time since i have done any math.

 

[Mentallic]

It's actually y=a(x-4)(x+4) where a is some constant multiplier. Notice that this still makes it a quadratic with the specifications that its roots be 4 and -4.

Now what value of a will make the quadratic pass through y-axis at 1? Notice that this point is (0,1).

 

[cgaday]

-1/16th correct? - 1/16 * (x^2 -16) going to give you 1 at the y intercept.

would this be the actual curve function? -1/16X^2 + 1?

When I integrate that for the area, why is the value negative? this was what i received after integration when b = 8 and a = 0. answer -2.667

 

[Mentallic]

Yes that's correct. But you could've also found the value of a by plugging in y=1 and x=0 into y=a(x-4)(x+4) and solving for a.

By b and a do you mean the limits of integration \int_a^b ? If so, you're getting a negative answer because there is more area under the x-axis between 4\leq x \leq 8 than there is above the x-axis at 0\leq x\leq 4.

 

[cgaday]

yes those are my limits. with my sketch, the curve is above the x-axis from 0 to 8. How is there more area under the x-axis when I thought the area I was trying to get is above the x axis. Does that make any sense?

 

[Mentallic]

Weren't we just previously finding a quadratic that has roots of -4 and 4, and cuts the y-axis at 1? What does this tell you about where the quadratic is above the x-axis, and it's not between 0 and 8.

 

[cgaday]

i'm sorry, I just was taking the difference between the two to have a total of 8. Yes that is correct.

back to my previous question regarding why the area is negative when I thought all the integral did was deteremine the area above the x axis. and between the curve.

 

[Mentallic]

Well just because the difference between -4 and 4 is 8, doesn't mean you integrate from 0 to 8. Integrating from 0 to 8 is asking for the area between the x-axis and the curve in between x=0 and x=8, but wherever the curve is under the x-axis, this is taken as negative area.

The integral does exactly that, but the curve isn't above the x-axis for all of the values of x between 0 and 8, as per my previous post.

 

[Mark44]

would this be the actual curve function? -1/16X^2 + 1?

You keep omitting one side of the equation that defines the relationship between x and y. It should be [color = red]y =[/color] (-1/16)x² + 1

When I integrate that for the area, why is the value negative? this was what i received after integration when b = 8 and a = 0. answer -2.667

What area are you trying to find? You haven't given a description of it. If you are trying to find the area between the parabola and the x-axis, you should not get a negative value for the area. In fact, you should never get a negative value for an area. If you do, that means you set up the integral incorrectly.

 

[rbmwz1]

Hey,

I hope this helps, I integrated your function.

Integrate (-1/16x^2 +1), from -4,4. = -1/48x^3+x, from -4, 4.

So. [(-1/48)(4^3)+4]-[(-1/48)(-4^3) - 4]

= [-4/3 +4 ] - [4/3 -4]
= 8-(8/3)
= 16/3

This should give you the area under the graph from -4 --> 4
hope this works. Let me know if it still gives you a problem.

 

[cgaday]

thank you. I see where I went wrong. I was still using limits of 0 and 8. Thanks for the help.

Ok. now another question relating to the same topic. What if the y-intercept is negative. Then would you get a negative area?

These are my new variables. x=-4.5 and x = 4.5
y-intercept = - .25

y = 1/81x^2 - .25


Integrate (1/81x^2 +1), from -4.5,4.5 = 1/243x^3+x, from -4.5, 4.5

So. [(1/243)(4.5^3)-4.5]-[(1/243)(-4.5^3) + 4.5]

= [.375 -4.5 ] - [-.375 + 4.5]
= -8.25



when I do this by hand I get -8.25, when I check with my calculator I achieve -1.5 Did I do something wrong?

 

[Mentallic]

Why did you integrate 1/81*x²+1 when you need to integrate 1/81*x²-1/4

 

[Mark44]

What if the y-intercept is negative. Then would you get a negative area?

You should never get a negative area. If you get a negative number for the area, you have set up the integral incorrectly.

For example, if the problem is to find the area between the graph of y = x² - 2x and the x-axis, the integral for the area is
\int_0^2 [0 - (x^2 - 2x)]~dx = \int_0^2 - x^2 + 2x~dx = \left.\frac{-x^3}{3} + x^2\right|_0^2
= -8/3 + 4 = 4/3

If you naively set up this integral as
\int_0^2 x^2 - 2x~dx
you will get -4/3, which is not the area.

The graph of y = x² - 2x has intercepts at x = 0 and x = 2, and lies below the x-axis between these intercepts. The typical area element has an area of (0 - ( x² - 2x) \Delta x, or ( -x² + 2x) \Delta x.

 

[cgaday]

I did. that was a typo.

For Mark:

So even though the line is y = x^2 - 2x

you changed the way it was in the integral so that x^2 in negative, and 2x is positive? I'm confused, why would you do that.
Like you demonstrated you will get a negative area, when integrating your origional formula.

So all I need to do is integrate my curve function after I multiply it by a negative.

 

[Mentallic]

If you have some function y, when you graph -y everything is flipped in the x-axis. This means when you take -(x²-2x), the graph is flipped such that the roots are the same, but the curve between the roots is now above the x-axis and thus the integral gives you a positive value for the area.

You could also take the absolute value of the integral, so you don't have to worry about whether the graph lies below or above the x-axis. (But be warned, if the graph lies above and below in some region, the area above the x-axis will cancel out the "negative area" below the axis).

 

[Mark44]

No, I didn't multiply these by -1. I made sure that the dimensions of the rectangle I was using had positive lengths. To do this, I subtract the smaller number from the larger number. On the interval [0, 2] the graph of y = x^2 - 2x lies below the x-axis. If I want the distance from a point (x, y) to the point directly above it on the x-axis, y - 0 will give me a negative value. 0 - y = 0 - (x^2 - 2x) will give me a positive value.

For example, (1, -1) is a point on this curve. This point is 0 - (-1) = 1 unit away from the x-axis. That's all there is to it.

 

[cgaday]

I have been working on this equation off and on but nothing seems to be correct.

I have x intercepts of -4.5 and 4.5
The y-intercept is .020833

(x-4.5)(x+4.5)

x^2 - 20.25
thus giving (-1/81x^2 + .020833)
the equation I have developed is (-1/81x^2 + .020833)

i still received a negative area of -0.562503.

Is this due to their being more area on the opposite side of the x-axis?

 

[Mark44]

I have been working on this equation off and on but nothing seems to be correct.

I have x intercepts of -4.5 and 4.5
The y-intercept is .020833

(x-4.5)(x+4.5)

Again, you are leaving off one side of the equation. By not working with equations, you are making things much harder for yourself than they need to be. If the equation is y = (x-4.5)(x+4.5), you don't get a y-intercept of .020833.

If you are given that the x-intercepts of a quadratic function are 4.5 and -4.5, the equation will be y = a(x-4.5)(x+4.5). Since the y-intercept is .020833, the point (0, .020833) must satisfy the equation y = a(x-4.5)(x+4.5). Use this point and solve for a.

x^2 - 20.25
thus giving (-1/81x^2 + .020833)
the equation I have developed is (-1/81x^2 + .020833)

This is NOT an equation. An equation has an = symbol in it.

i still received a negative area of -0.562503.

You should not get a negative area. Please reread what I said in post #15.

Is this due to their being more area on the opposite side of the x-axis?

 

[Mentallic]

I have a feeling that you're still taking the limits of integration from 0 to 8 for your quadratic with roots -4 and 4. Just because the distance between the roots is 8, doesn't mean you take your limits from 0 to 8. What this is does is as follows:

http://img10.imageshack.us/img10/9609/parabolaintegration1.png

Uploaded with ImageShack.us

Notice that it takes the area between the curve and the x-axis from x=0 to x=8 (these are your limits of integration, \int_0^8). The area above the x-axis, from x=0 to x=4, is positive area, while the area below the x-axis, is considered negative area (of course area is always positive, but in integration, this comes out as negative).
Now since there is slightly more area underneath than above the x-axis, your answer will turn out slightly negative which tells you nothing important. Say you wanted to find the total area of the picture shown above from x=0 to x=8, you need to split the integral into x=0 to x=4, and then take the negative of x=4 to x=8.

But what I think you want is this:

http://img704.imageshack.us/img704/1179/parabolaintegration2.png

Uploaded with ImageShack.us

Change your limits accordingly to find the area between the parabola and the curve.

 

[cgaday]

I like the graphs. I have been using the correct limits lately. Thanks for the confirmation.

Mark:

I am trying to understand the correlation between the y intercept and the limits.

I presume that you set .020833 = a(x-4.5)(x+4.5)

so i achieved a = .020833/((x-4.5)(x+4.5))

a = .020833/(x^2-20.25)

I'm lost.

 

[Mentallic]

Since your quadratic has roots of x=\pm 4.5 then it can be transformed into y=a(x-4.5)(x+4.5)
a is some constant multiplier. If we used a=1 then we would have the quadratic y=x^2-20.25 but this cuts the y-axis at -20.25 and that doesn't satisfy your other condition that the y-intercept is .020833

So we need to find the correct a, and we know it must be of that form above (in order to satisfy the condition of the two roots and that it is a quadratic). We know what point does satisfy the quadratic - besides (4.5 , 0) and (-4.5 , 0) - and that is (0 , 0.20833).

If we substitute this point into our quadratic, we can find the value of a. That is, sub x=0 and y=0.020833 into the equation.

a=0.020833/(-20.25)

The correlation between the y-intercept and the limits is that while you keep your limits the same (whatever the roots are, in this case -4.5 to 4.5), you need to find the correct quadratic to integrate. We've done this above.

 

[cgaday]

.020833/20.25 = .001029 ?

-.001029x^2 + .020833

integrated I received .125001 doesn't seem correct. I think I'm making this too difficult, or my equation is just messing with my head.

 

[Mentallic]

No, what was the original function again?

y=a(x^2-0.020833)

Now sub your value of a into that.

 

[cgaday]

I thought this was the origional function. y = x^2 - 20.25

with a y-intercept at .25



is this correct?
y = -0.001029x^2 + .000021

 

[Mark44]

I thought this was the origional function. y = x^2 - 20.25

with a y-intercept at .25

No, not at all. This function has x-intercepts of 4.5 and -4.5, but the y-intercept is -20.25, not .25.

There is nothing very complicated about the y-intercept. It is the y-value when x = 0. And the x-intercepts are the x-values that correspond to a y-value of 0.

is this correct?
y = -0.001029x^2 + .000021

It is not clear to me what exactly is the given information in this problem. I'm pretty sure the x-intercepts are 4.5 and -4.5 and that you're working with a quadratic, but it seems to me that the y-intercept has changed from .25 to .020833. Is this two different problems?

You are apparently trying to find the area bounded by the graph of the quadratic and the x-axis, which is why you are integrating from -4.5 to 4.5. However, you can't do the integration until you find the formula for the function to be integrated.

So what is the y-intercept?

 

[cgaday]

the y intercept is .020833. the reason it was .25 is because that was in inches. and the x intercepts were in feet. So i divided .25 by 12 to get everything in the same units.

Sorry i have been so confusing. So are we on the correct path?

 

[Mark44]

OK, now I see what you're doing. It's probably best to leave the numbers as fractions, rather than using decimals, especially since you are going to integrate. Doing so gives
y = -(1/972)(x^2 - 81/4).
It's just as well to leave it in this form or as y = -(1/972)(x^2 - 20.25).

Now integrate.
\frac{-1}{972}\int_{-4.5}^{4.5} x^2 - 20.25 dx

 

[cgaday]

Thank you soo much.

how did you convert from decimal to fractions. Those are easier to deal with.

I have an ti-89 but didn't see how to go to fractions.

 

[Mark44]

Mostly, I left things as fractions, writing 4.5 as 9/2, for example, so 4.5^2 = 81/4. Also, 1/4 inch = 1/48 ft.