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

**moskowfreak1111** · Apr 12th 2013, 12:34 PM

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?

**HallsofIvy** · Apr 12th 2013, 04:34 PM

There is no simple "algebraic" method. You say you aren't supposed to solve it "graphically". Is some other numerical method allowed?

**moskowfreak1111** · Apr 12th 2013, 06:53 PM

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?

**x3bnm** · May 15th 2013, 01:42 PM

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: $x = -3.16715$

p_n-1	p_n-2	f(p_n-1)	f(p_n-2)	p_n
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.

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