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Thread: How to find points of intersection of a parabola and an exponential w/o graphing?

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    How to find points of intersection of a parabola and an exponential w/o graphing?

    Hey mathhelpforum.com, this is my first post! And I need some help with math!

    The equation is 3^x+5=x^2-5, we aren't supposed to be able to solve them yet non-graphically, so I wonder if it is even possible, if so, how?
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    Re: How to find points of intersection of a parabola and an exponential w/o graphing?

    There is no simple "algebraic" method. You say you aren't supposed to solve it "graphically". Is some other numerical method allowed?
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    Re: How to find points of intersection of a parabola and an exponential w/o graphing?

    Well, we can, and that's the only way to solve it is graphically my teacher says, but I don't want to, I want to actually learn how to do it, what would be the non-simple ways to do it?
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    Senior Member x3bnm's Avatar
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    Re: How to find points of intersection of a parabola and an exponential w/o graphing?

    The equation is 3^x+5=x^2-5, we aren't supposed to be able to solve them yet non-graphically, so I wonder if it is even possible, if so, how?
    For Nonlinear equations like yours you can use secant method.

    Explanation of Secant Method:

    Suppose there's a nonlinear function $\displaystyle y = f(x)$ in the image given below:

    How to find points of intersection of a parabola and an exponential w/o graphing?-secant.png

    (a) Now take any two values of $\displaystyle x$ along the $\displaystyle X$ axis Say the first one is $\displaystyle p_0$ and the second one is $\displaystyle p_1$

    (b) plug the value of $\displaystyle p_0$ and $\displaystyle p_1$ into $\displaystyle y = f(x)$ and solve for $\displaystyle y$.

    (c) You will get two points. They are $\displaystyle (p_0, f(p_0))$ and $\displaystyle (p_1, f(p_1))$

    (d) draw a secant line that goes through $\displaystyle (p_0, f(p_0))$ and $\displaystyle (p_1, f(p_1))$

    That secant line is $\displaystyle L1$

    $\displaystyle L1$ intersects $\displaystyle X$ axis at $\displaystyle p_2$

    (e) If you look carefully we can now write:

    $\displaystyle \frac{f(p_1)}{p_1 - p_2} = \frac{f(p_1)-f(p_0)}{p_1 - p_0}......[\text{Equation (1)}]$

    (f) Plug the values of $\displaystyle p_0$, $\displaystyle f(p_0)$, $\displaystyle p_1$ and $\displaystyle f(p_1)$ and you will get the value $\displaystyle p_2$

    (g) Now plug the value of $\displaystyle p_2$ into $\displaystyle y = f(x)$ as the value of $\displaystyle x$ and solve for $\displaystyle y$

    you'll get the point $\displaystyle (p_2, f(p_2))$

    (h) A secant line can be drawn from $\displaystyle (p_1, f(p_1))$ to the point $\displaystyle (p_2, f(p_2))$ which is $\displaystyle L2$.

    $\displaystyle L2$ intersects $\displaystyle X$ axis at $\displaystyle p_3$

    (i) Again we can write:

    $\displaystyle \frac{f(p_2)}{p_2 - p_3} = \frac{f(p_2)-f(p_1)}{p_2 - p_1}$

    Now plug in $\displaystyle p_1$, $\displaystyle f(p_1)$, $\displaystyle p_2$ and $\displaystyle f(p_2)$ and solve for $\displaystyle p_3$

    (j) Again plug this value of $\displaystyle x = p_3$ into $\displaystyle y =f(x)$ and find $\displaystyle f(p_3)$

    (k) Now continue this process iteratively until a value of $\displaystyle x$ is reached that will make $\displaystyle y = f(x)$ closer to zero.
    And that's it that's the root of the equation.

    We can write Equation(1) as:

    $\displaystyle \frac{f(p_{n-1})}{p_{n-1} - p_n} = \frac{f(p_{n-1})-f(p_{n-2})}{p_{n-1} - p_{n-2}}......[\text{Equation (2)}]$


    Solving your question:

    You want to find the root of the equation is:

    $\displaystyle 3^x - x^2 + 10 = 0$

    So the function is:

    $\displaystyle f(x) = 3^x - x^2 + 10.......(3)$

    (1) Take any two $\displaystyle x$ values along the $\displaystyle X$ axis. Say the first point $\displaystyle p_{n-2} = 2$ and $\displaystyle p_{n-1} = 3$

    $\displaystyle f(p_{n-2}) = f(2) = 15.....\text{[by plugging in } x = 2 \text{into function labeled as (3)]}$

    And:

    $\displaystyle f(p_{n-1}) = f(3) = 28$

    (2) Now plug in $\displaystyle p_{n-1}, f(p_{n-1}), p_{n-2}, f(p_{n-2})$ into Equation(2) and solve for $\displaystyle p_{n}$

    You'll get $\displaystyle p_n = 0.84615$

    (3) Now let $\displaystyle p_{n-2} = p_{n-1} = 3$ and $\displaystyle p_{n-1} = p_n = 0.84615$

    And do step (1) to (3) then again and again until plugging in the new found $\displaystyle p_n$ in each step as value of $\displaystyle x$ into $\displaystyle y = 3^x - x^2 + 10$ gets reasonably closer to zero.


    After 10 iterations I got the root of $\displaystyle y = 3^x - x^2 + 10$. And it's $\displaystyle x = -3.16715$

    If I plug this value in $\displaystyle y = 3^x - x^2 + 10$ I get $\displaystyle -0.00002$ which is reasonably close.

    So the approximate answer is: $\displaystyle x = -3.16715$

    pn-1
    pn-2 f(pn-1) f(pn-2) pn
    3 2 28 15 0.84615
    0.84615 3 11.81750 28 -0.72673
    -0.72673 0.84615 9.92191 11.81750 -8.95951
    -8.95951 -0.72673 -70.27277 9.92191 -1.74531
    -1.74531 -8.95951 7.10088 -70.27277 -2.40739
    -2.40739 -1.74531 4.27549 7.10088 -3.40928
    -3.40928 -2.40739 -1.59957 4.27559 -3.13651
    -3.13651 -3.40928 0.19418 -1.59957 -3.16604
    -3.16604 -3.13651 0.00705 0.19418 -3.16715
    -3.16715
    -3.16604 -0.00002 0.00705 -3.16715

    Hope this answers your question.
    Last edited by x3bnm; May 15th 2013 at 02:34 PM.
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