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I know this seems completely amateur, but for whatever reason I cannot solve this.

I need to find the intersection values for $y= 10-0.00001x^2$ and $y=5+0.005x$

Any help would be much appreciated, thanks

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Functions do not intersect. What you seem to need is to find the coordinates of the points of intersection of the graphs of those two functions. –  Mariano Suárez-Alvarez Apr 19 '12 at 18:14
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4 Answers 4

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The points of intersection must satisfy both equations. Therefore if ($x_i$,$y_i$) is a solution, then it must be true that $y_i = 10-0.0001(x_i)^2$, and also that $y_i = 5 + 0.005x_i$.

So if ($x_i,y_i$) is a solution, then we must have $y_i = 10-0.0001(x_1)^2 = 5 + 0.005x_i$. (*)

Solve the quadratic labelled (*), and you get two possible pairs ($x_1$, $y_1$), ($x_2$, $y_2$). Now you should check that these solutions really work by plugging both of them back into the original equation (which you were originally trying to solve).

After a quick check, we see that they are both solutions to the original equation.

Are there any other solutions? No, because if ($x_3$, $y_3$) is another solution, then ($x_3$, $y_3$) does not satisfy the quadratic equation values of x solutions which satisfy equation (*) (due to the fact that a quadratic equation has at most 2 solutions), and so a 3rd solution does not exist.

From now on don't have to do this every time, and instead of writing out ($x_i$, $y_i$) every time, we can just solve the quadratic because now we know that doing so gives all the solutions. But you can use a similar method in your head when asked to find all the solutions to a pair of equations.

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I edited your answer in order to use inline Latex whenever you seemed to need a mathematical symbol or equation. Just add the $ symbol around the part of the sentence that has to be converted to math. –  logc Feb 26 at 13:51
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The intersection is precisely the points where $$10 - 0.00001x^2= 5+0.005x.$$ This corresponds to a quadratic equation on $x$, $$0.00001x^2 + 0.005x - 5 = 0$$ or equivalently, $$x^2 + 500x - 500000 = 0.$$ which can be solved by the usual methods. Once you know the (up to) two values of $x$, they give you the corresponding values of $y$.

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We have $y=10-0.00001x^2$ and $y=5+0.005x$. This is true if and only if $10-0.00001x^2=5+0.005x$, which simplifies to $$0.00001x^2+0.005x-5=0.\tag{$\ast$}$$ It might be nice to multiply through by $10^5$. We get $$x^2+500x-500000=0.\tag{$\ast\ast$}$$ No big improvement!

Now use the ordinary Quadratic Formula. If $a \ne 0$, then the roots of $ax^2+bx+c=0$ are $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$.

Now you can compute exact expressions for the roots. Or if you just want them to high accuracy, the calculator handles things quite nicely. We can use the Quadratic Formula on $(\ast\ast)$, but also directly on $(\ast)$.

Remark: By changing the numbers somewhat, we can produce interesting calculational problems. For instance, if you want to find the roots of $x^2-10^8 x+1$ to say $4$ significant figures, naive use of the Quadratic Formula and a simple scientific calculator will lead us to conclude that the "small" root is $0$, which is not correct to even one significant figure. The reason there is a problem is roundoff error in the calculator. But that issue does not come up with the numbers in your quadratic.

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Set the two expressions equal to each other: $$10-.00001x^2=y=5+0.005x$$ $$.00001x^2+.005x-10+5=0$$ $$.00001x^2+.005x-5=0$$ $$x^2+500x-500000=0$$ Now apply the quadratic formula:$$x=\frac{-500\pm\sqrt{250000+2000000}}{2}$$ $$x=\frac{-500\pm1500}{2}$$ $$x=-250\pm750$$

To find the corresponding $y$ values, plug in $500$ and $-1000$ for $x$ in the original equations.

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thank you very much, I see how to do it now :) except its 5+0.005x, nonetheless I see! –  Sam Creamer Apr 19 '12 at 17:11
    
Yeah, typo there, fortunately I didn't carry it over to the next line. Modulo any more careless sign or order-of-magnitude errors, it should be correct as now shown (you should repeat the calculation on your own to make sure). –  Brett Frankel Apr 19 '12 at 17:14
    
@RossMillikan Now corrected, and the final answers check out in the original equation. –  Brett Frankel Apr 19 '12 at 21:39
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