Need help finding intercept in a polynomial function
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Hello,
I have a polynomial function and I would like to find all the "x" values for a "y" value.
For example I have tried to plot this function:
x=[0:0.05:23]
f=@(x) 152.094827571469-206.917405701416*x+188.571689289441*(x.^2)-93.85238987*(x.^3)+25.7783269818991*(x.^4)-4.34657779503011*(x.^5)+0.485315204*(x.^6)-0.0353709647924109*(x.^7)+0.00158130195601334*(x.^8)-0.0000388214150997567*(x.^9)+0.000000397986046326228*(x.^10)
y=f(x)
plot(x,y)
grid on
I have tried to do all the intercepts with the interp1 formula (for example when y=600)
x0=interp1(y,x,600,'nearest')
x1=interp1(y,x,600,'linear')
x2=interp1(y,x,600,'spline')
x3=interp1(y,x,600,'pchip')
x4=interp1(y,x,600,'cubic')
x5=interp1(y,x,600,'v5cubic')
As you can see in the graph I should get 2 points when y=600 but I only one and it is the same for all the y values I use.
Please if you know any way to find all the interceptions when y=600 or any value it would be very helpful!
Thank you for your time and help.
采纳的回答
James Tursa
2019-11-27
编辑：James Tursa
2019-11-27
0 个投票
Why can't you just find the real roots of f(x)-600? What am I missing here?
2 个评论
Guillaume
2019-11-27
Indeed, and that would be trivial to do if the polynomial was expressed as a vector of coefficients instead of being hardcoded in an anonymous function:
p = fliplr([152.094827571469;
-206.917405701416;
188.571689289441;
-93.85238987;
25.7783269818991
-4.34657779503011
0.485315204
-0.0353709647924109
0.00158130195601334
-0.0000388214150997567
0.000000397986046326228].')
f = @(x) polyval(p, x);
r = roots(p - [zeros(1, 10), 600]);
r(imag(r) == 0)
Gerardo Flores Sempertegui
2019-11-28
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2019-11-27
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