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Find Asymptotes, Critical, and Inflection Points

This example describes how to analyze a simple function to find its asymptotes, maximum, minimum, and inflection point.

Define a Function

The function in this example is


First, create the function.

syms x
num = 3*x^2 + 6*x -1;
denom = x^2 + x - 3;
f = num/denom
f = 

3x2+6x-1x2+x-3(3*x^2 + 6*x - 1)/(x^2 + x - 3)

Plot the function by using fplot. The fplot function automatically shows vertical asymptotes.


Figure contains an axes. The axes contains an object of type functionline.

Find Asymptotes

To find the horizontal asymptote of f mathematically, take the limit of f as x approaches positive infinity.

ans = 3sym(3)

The limit as x approaches negative infinity is also 3. This result means the line y=3 is a horizontal asymptote to f.

To find the vertical asymptotes of f, set the denominator equal to 0 and solve it.

roots = solve(denom)
roots = 

(-132-12132-12)[- sqrt(sym(13))/2 - sym(1/2); sqrt(sym(13))/2 - sym(1/2)]

roots indicates that the vertical asymptotes are the lines




Find Maximum and Minimum

You can see from the graph that f has a local maximum between the points x=2 and x=0. It also has a local minimum between x=6 and x=2. To find the x-coordinates of the maximum and minimum, first take the derivative of f.

f1 = diff(f)
f1 = 

6x+6x2+x-3-2x+13x2+6x-1x2+x-32(6*x + 6)/(x^2 + x - 3) - ((2*x + 1)*(3*x^2 + 6*x - 1))/(x^2 + x - 3)^2

To simplify this expression, enter the following.

f1 = simplify(f1)
f1 = 

-3x2+16x+17x2+x-32-(3*x^2 + 16*x + 17)/(x^2 + x - 3)^2

Next, set the derivative equal to 0 and solve for the critical points.

crit_pts = solve(f1)
crit_pts = 

(-133-83133-83)[- sqrt(sym(13))/3 - sym(8/3); sqrt(sym(13))/3 - sym(8/3)]

As the graph of f shows, the function has a local minimum at


and a local maximum at


Plot the maximum and minimum of f.

hold on
plot(double(crit_pts), double(subs(f,crit_pts)),'ro')
title('Maximum and Minimum of f')
text(-4.8,5.5,'Local minimum')
text(-2,4,'Local maximum')
hold off

Figure contains an axes. The axes with title Maximum and Minimum of f contains 4 objects of type functionline, line, text.

Find Inflection Point

To find the inflection point of f, set the second derivative equal to 0 and solve for this condition.

f2 = diff(f1);
inflec_pt = solve(f2,'MaxDegree',3);
ans = 3×1 complex

  -5.2635 + 0.0000i
  -1.3682 - 0.8511i
  -1.3682 + 0.8511i

In this example, only the first element is a real number, so this is the only inflection point. MATLAB® does not always return the roots to an equation in the same order.

Instead of selecting the real root by indexing into inter_pt, identify the real root by determining which roots have a zero-valued imaginary part.

idx = imag(double(inflec_pt)) == 0;
inflec_pt = inflec_pt(idx)
inflec_pt = 

-13916954-2197181/3-16954-2197181/3-83- 13/(9*(sym(169/54) - sqrt(sym(2197))/18)^sym(1/3)) - (sym(169/54) - sqrt(sym(2197))/18)^sym(1/3) - sym(8/3)

Plot the inflection point. The extra argument [-9 6] in fplot extends the range of x values in the plot so that you can see the inflection point more clearly, as the figure shows.

fplot(f,[-9 6])
hold on
plot(double(inflec_pt), double(subs(f,inflec_pt)),'ro')
title('Inflection Point of f')
text(-7,1,'Inflection point')
hold off

Figure contains an axes. The axes with title Inflection Point of f contains 3 objects of type functionline, line, text.