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Sketching the Curve of a Polynomial Function

Let’s take the function f(x) = y = x3 – 6x2 + 11x – 6.

 

Being a polynomial of degree 3, there are two ways to sketch the graph for this function.

  1. By not solving the polynomial, but getting most parameters for sketching a curve
  2. By solving the polynomial, and getting an accurate set of parameters for sketching the curve

 

Method 1: Sketching the curve of a polynomial function without solving the function.

 

In this method, we’ll skip steps 1 to 4 of curve sketching and go straight to steps 5, 6 and 7. We can roughly sketch the graph with stationary point, point of inflection, and y-intercept.

Step 5: Find stationary points (put y’ = f'(x) = 0)

f'(x) = y’ = 3x2 – 12x + 11 = 0

Applying the quadratic equation x = {-b~pm~sqrt{b^2~-~4ac}}/{2a}

x = {12~pm~sqrt{12^2~-~4*3*11}}/{2*3}

{12~pm~sqrt{144~-~132}}/{6}

{12~pm~sqrt{12}}/{6}

= 2.58 or 1.42

When x = 2.58, y = -0.385

When x = 1.42, y = 0.385

So (2.58, -0.385) and (1.42, 0.385) are stationary points.

 

Step 6: Find the point of inflection

y” = f”(x) = 6x – 12

y” at x = 2.58 = 6×2.58 – 12 = 3.48 > 0

Since y” > 0, it is the local minimum point.

y” at x = 1.42 = 6×1.42 – 12 = -3.48 < 0

Since y” < 0, it is the local maximum point.

y” = f”(x) =  0    doubleright 6x – 12 = 0, or 6x = 12 or x = 2

When x = 2, y = (2)3 – 6(2)2 + 11×2 – 6 = 0

(2, 0) could therefore be a point of inflection.

 

Step 7: Check for concavity around x = 2.

x 1 2 3
f”(x) 6 – 12 0 18 – 12
f”(x) = -6 < 0 0 = 6 > 0

 

The concavity of the curve changes, hence (2, 0) is a point of inflection.

When x = 0, y = (0)3 – 6(0)2 + 11×0 – 6 = -6

 

Now we can sketch the curve with the following points –

  • maximum point right (1.42, 0.385)
  • minimum point  right (2.58, -0.385)
  • point of inflection right  (2, 0)
  • y-intercept right (0, -6)

To find some in-between points, we put x = 1 to get y = 0

Put x = 3 to get y = 0, and x = 4 to get y = 6.

curve sketching for a polynomial function

 

 

Method II: Sketching the curve of a polynomial function by solving the function

 

Step 1: Find the x-intercepts, by putting y = 0 and solving the equation

f(x) = y = x3 – 6x2 + 11x – 6 = 0

Put x = 1 (since alternate coefficients add to the same) in the equation.

(1)3 – 6(1)2 + 11×1 – 6 = 1 – 6 + 11 – 6 = 0

So (x – 1) is a solution, and we can divide the function by (x – 1) to get (x2 – 5x + 6)

x3 – 6x2 + 11x – 6 = (x – 1)(x2 – 5x + 6)

= (x -1)(x – 2)(x – 3)

x-intercepts are (1, 0), (2, 0), and (3, 0).

 

Step 2: Find the y-intercepts, by putting x = 0

y = (0)3 – 6(0)2 + 11×0 – 6 = 0 – 0 + 0 – 6 = -6

So y-intercept point is (0, -6)

 

Step 3: Check if the curve is symmetric, i.e. is the function odd or even.

f(x) = x3 – 6x2 + 11x – 6

f(-x) = (-x)3 – 6(-x)2 + 11x – 6 =   -x3 – 6x2 – 11x – 6

=  -(x3 + 6x2 + 11x + 6)

This is neither an odd nor an even function. So there is no symmetry.

 

Step 4: Check for any discontinuities, and find the asymptotes, if any, or the limits

When {x}~right~infty{y}~right~infty, since {infty}^3 = large value.

{x}~right~-infty{y}~right~-infty, since {infty}^-3 = very small value.

There are no discontinuities or asymptotes to find for this function.

 

Steps 5, 6 and 7 are exactly the same as the Method 1 described above. Using all the points, we can now plot the graph.

curve sketching for a polynomial function