LEAVE A COMMENT FOR US

Home > Calculus > Curve Sketching > Sketching Curves with Asymptotes – Example 1

Sketching Curves with Asymptotes – Example 1

We earlier saw how to sketch the curve of a function and a polynomial function (with and without solving the polynomial function).

We now look at an example of sketching curves with asymptotes, i.e. how to sketch a curve that has asymptotes.

Example 1: Sketch a curve for f(x) = 2/{x^2~+~3}

Step 1: Find the y-intercepts, when x = 0

y = 2/3

Therefore (0, 2/3) is the y-intercept

 

Step 2: We cannot find the x-intercepts, since y~<>~0

 

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

f(x) =  2/{x^2~+~3}

f(-x) =  2/{x^2~+~3}

f(x) = f(-x)

So this is an even function, and is symmetric about y-axis.

 

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

lim{x right infty}~{2/{x^2~+~3}}  =  lim{x right infty}~{{2/x^2}/{1~+~{3/x^2}}}

= 0/{1~+~0}  =  0+

lim{x right - infty}~{2/{x^2~+~3}}  =  lim{x right - infty}~{{2/x^2}/{1~+~{3/x^2}}}

=  0-

 

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

dy/dx  =  ~-2/({(x^2~+~3)}^2}~.2x  =  ~{-4x}/({(x^2~+~3)}^2}

dy/dx  =  0   So ~{-4x}/({(x^2~+~3)}^2}  = 0

doubleright  x  =  0

(0,~2/3) is a stationary point.

 

Step 6: Find the point of inflection

{d^2y}/{dx^2}  =  ~{(-4){(x^2~+~3)^2}~+~4x(2)(x^3~+~3).2x}/({(x^2~+~3)}^4}

=  {~4(x^2~+~3)({-(x^3~+~3)}~+~16x^2)}/({(x^2~+~3)}^4}

=  {~4(15x^2~-~3)}/({(x^2~+~3)}^3}

For point of inflection, {d^2y}/{dx^2}  = 0.

Therefore 15x2 – 3 = 0

x2  = 1/5, or  x  =  pm{1/sqrt{5}}

When x = 0, {d^2y}/{dx^2} =  {~4(15(0)^2~-~3)}/({((0)^2~+~3)}^3}  =  {~4(-3)}/{3}^3  = -4/9  < 0

Hence (0,~2/3) is a maximum point.

Test for point of inflection at x =  1/sqrt{5}  = 0.447

x 0.4 1/sqrt{5} = 0.447 0.5
f”(x) -0.07 < 0 0 0.087 > 0

When x = 0.447, y = 0.625

So (0.447, 0.625) is a point of inflection.

 

Test for point of inflection at x =  -1/sqrt{5}  = -0.447

x -0.5 -1/sqrt{5} = -0.447 -0.4
f”(x) 0.087 > 0 0 -0.07 < 0

When x = -0.447, y = 0.625

So (-0.447, 0.625) is a point of inflection.

 

curve sketching with asymptotes

See also other examples of sketching curves with asymptotes –

Sketching curves with asymptotes – Example 2

Sketching curves with asymptotes – Example 3

Sketching curves with asymptotes – Example 4