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Math Help - 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 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 x along the X axis Say the first one is p_0 and the second one is p_1

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

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

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

    That secant line is L1

    L1 intersects X axis at p_2

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

    \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 p_0, f(p_0), p_1 and f(p_1) and you will get the value p_2

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

    you'll get the point (p_2, f(p_2))

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

    L2 intersects X axis at p_3

    (i) Again we can write:

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

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

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

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

    We can write Equation(1) as:

    \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:

    3^x - x^2 + 10 = 0

    So the function is:

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

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

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

    And:

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

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

    You'll get p_n = 0.84615

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

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


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

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

    So the approximate answer is: 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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