Informative line

Curve Sketching

Learn how to find horizontal asymptotes using limits and vertical asymptotes of rational functions, practice graphing using first and second derivatives, curve sketching calculus problems.

Horizontal Asymptote

• A line $$y=L$$ is called the horizontal asymptote of a curve $$y=f(x)$$ if either
1. $$\lim \limits_{x\to\infty} f(x) = L$$
2. $$\lim \limits_{x\to-\infty} f(x)= L$$  • From the graph  $$\lim \limits_{x\to\infty} f(x)=1$$    or $$\lim \limits_{x\to-\infty} f(x)= -3$$ that is  lines $$y=1$$ and $$y=-3$$ are both horizontal asymptote.

The horizontal asymptotes of the curve   $$y=f(x)$$  as shown below is /are

A $$y=1 \,\,\,{\text{&}} \,\,\,y=2$$

B $$y=0 \,\,\,{\text{&}} \,\,\,y=3$$

C $$y=0\,\,\,{\text{&}} \,\,\,y=2$$

D $$y=3 \,\,\,{\text{&}} \,\,\,y=1$$

×

Horizontal asymptotes are $$y=\lim \limits_{x\to\infty} f(x)$$ or $$\lim \limits_{x\to-\infty} f(x)$$ . $$\lim \limits_{x\to\infty} f(x) =1$$   and $$\lim \limits_{x\to-\infty} f(x)=2$$ $$\therefore$$ Horizontal asymptotes are  $$y=1 \,\,\,{\text{&}} \,\,\,y=2$$ . The horizontal asymptotes of the curve   $$y=f(x)$$  as shown below is /are A

$$y=1 \,\,\,{\text{&}} \,\,\,y=2$$

.

B

$$y=0 \,\,\,{\text{&}} \,\,\,y=3$$

C

$$y=0\,\,\,{\text{&}} \,\,\,y=2$$

D

$$y=3 \,\,\,{\text{&}} \,\,\,y=1$$

Option A is Correct

Finding the Horizontal Asymptote of a function by Calculating Limits at positive and negative infinity

• To calculate the horizontal asymptote of any curve $$y=f(x)$$ , Find
1. $$\lim\limits_{x\to\infty} f(x) = L_1$$
2. $$y=\lim\limits_{x\to-\infty} f(x) = L_2$$

then $$y=L_1\,{\text {&}} \,\,y= L_2$$ are called horizontal asymptotes.

Find the horizontal asymptotes of the curve $$y = \dfrac{\sqrt{4x^2+3}}{5x+1}$$ .

A $$y= \dfrac{2}{5}$$ and $$y= \dfrac{-2}{5}$$

B $$y= 0$$ and $$y= 2$$

C $$y= \dfrac{4}{5}$$  and $$y= \dfrac{-4}{5}$$

D $$y= 5$$ and  $$y= -4$$

×

$$\lim\limits_{x\to\infty} f(x) =\lim\limits_{x\to\infty} \dfrac{\sqrt{4x^2+3}}{5x+1}$$

Divide numerator and denominator by $$x$$

$$\Rightarrow \lim\limits_{x\to\infty} \left(\dfrac{\dfrac{\sqrt{4x^2+3}}{x}}{\dfrac{5x+1}{x}}\right)$$

$$\Rightarrow \lim\limits_{x\to\infty} \left(\dfrac{{\sqrt{4+\dfrac{3}{x^2}}}}{{5+\dfrac{1}{x}}}\right)$$

$$\dfrac{\sqrt{\lim\limits_{x\to\infty}\,4+\lim\limits_{x\to\infty}\,\dfrac{3}{x^2}}}{\left(\lim\limits_{x\to\infty}\,5\right)+\left(\lim\limits_{x\to\infty}\dfrac{1}{x}\right)} = \dfrac{\sqrt{4+0}}{5+0}$$

$$= \dfrac{2}{5}$$

$$\therefore y= \dfrac{2}{5}$$ is one horizontal asymptote.

$$\lim\limits_{x\to-\infty} f(x) = \lim\limits_{x\to-\infty} \left(\dfrac{\sqrt{4x^2+3}}{5x+1}\right)$$

Divide numerator and denominator by $$x$$

$$\Rightarrow \lim\limits_{x\to-\infty} \left(\dfrac{\dfrac{\sqrt{4x^2+3}}{x}}{\dfrac{5x+1}{x}}\right)$$

$$\Rightarrow \lim\limits_{x\to-\infty} \left(\dfrac{{\dfrac{-\sqrt{4x^2+3}}{x^2}}}{{5+\dfrac{1}{x}}}\right) \left(\sqrt x^2 = |x|=-x\right)$$

$$\Rightarrow -\lim\limits_{x\to-\infty}\dfrac{\sqrt{4+\dfrac{3}{x^2}}}{5+\dfrac{1}{x}}= \dfrac{-\sqrt{\lim\limits_{x\to\infty}\,4+\lim\limits_{x\to\infty}\,\dfrac{3}{x^2}}}{\left(\lim\limits_{x\to\infty}\,5\right)+\left(\lim\limits_{x\to\infty}\dfrac{1}{x}\right)} = \dfrac{-\sqrt{4+0}}{5+0}$$

$$= \dfrac{-2}{5}$$

$$\therefore y= \dfrac{-2}{5}$$ is the other horizontal asymptote.

Find the horizontal asymptotes of the curve $$y = \dfrac{\sqrt{4x^2+3}}{5x+1}$$ .

A

$$y= \dfrac{2}{5}$$ and $$y= \dfrac{-2}{5}$$

.

B

$$y= 0$$ and $$y= 2$$

C

$$y= \dfrac{4}{5}$$  and $$y= \dfrac{-4}{5}$$

D

$$y= 5$$ and  $$y= -4$$

Option A is Correct

Sketching the Graph of a Polynomial (without using Derivatives)

• By using limits at infinity and $$x$$ and $$y$$ intercept (i.e. the point at which graph intersect  $$x$$ and $$y$$ axis ) we can get an idea of graph of a polynomial function.

Which of the following can be a rough  sketch of $$y= (2-x)(x+1)^4 (1-x)^6$$

A B C D ×

Find  $$x$$ and $$y$$ intercepts first.

$$y$$ intercept = $$f(0) = 2× 1^4 × 1^6 = 2$$

$$\therefore$$ graph should pass through (0,2)

To get $$x$$  intercept put $$y=0$$

$$\Rightarrow (2-x)(1+x)^4 (1-x)^6 = 0$$

$$\Rightarrow x=2,-1,+1 \to$$ graph passes through  $$(2,0), (1,0), (-1,0)$$

 Interval $$2-x$$ $$(1-x)^6$$ $$(1+x)^4$$ $$f$$ $$x >2$$ $$-$$ $$+$$ $$+$$ $$-$$ $$1 Look at the sign of \('f'$$ , it is negative in $$x >2$$ and positive every where.

$$=\lim\limits _{x\to\infty} \underbrace{(2-x)} _{-\infty} \underbrace{(x+1)^4}_{\infty} \underbrace{(1-x)^6}_{\infty} = -\infty$$

$$\lim\limits _{x\to-\infty} \underbrace{(2-x)} _{\infty} \underbrace{(x+1)^4}_{\infty} \underbrace{(1-x)^6}_{\infty} = \infty$$

Graph should be very high when $$x \to-\infty$$  and very low when $$x \to\infty$$

$$\therefore$$ Option $$(a)$$ is correct.

Which of the following can be a rough  sketch of $$y= (2-x)(x+1)^4 (1-x)^6$$

A B C D Option A is Correct

Sketching the Graph of a Function with some Desired Properties

• If we desire to sketch  the graph of a function which satisfies certain properties about the derivatives . we make use of the following points .
1. If $$f' (x)= 0$$ then there is a extreme  at that points .  2. If $$f' (x)> 0$$ in an interval then $$f$$ is increasing in the interval.

3.  If $$f' (x)< 0$$ in an interval then $$f$$ is decreasing in the interval.

4. If $$f'' (x)> 0$$ then $$f$$ is concave upwards .

5. If $$f'' (x)<0$$ then $$f$$ is concave downwards .

6. $$f(-x) = f(x) \Rightarrow$$ Symmetry about $$y$$ axis.

Sketch the graph of a function $$f$$ which satisfy the following condition . $$f'(3)=0, \,\,f(3) =-2, \,\,f(0) =0$$ $$f'(x)<0$$ if  $$0<x<3$$  $$\text {&}$$  $$f'(x)>0$$  if $$x>3$$ $$f''(x) <0$$ if $$0\leq x<2 \,\,\,or \,\,\,x>5 \,\,\,\, {\text {&}} \,\,\,\, f''(x) >0 \,\,if\,\,\,2<x<5$$ $$\lim\limits_{x\to\infty} f(x) =2$$ $$f(-x) = f (x) \forall x$$

A B C D ×

$$f(-x) = f(x) \Rightarrow$$ Graph is symmetric about $$y$$ axis .

$$\therefore$$ We sketch the graph to right of $$y$$ axis and then take reflection in $$y$$ axis .

$$f' (3) = 0 \Rightarrow$$ critical number at  $$x = 3$$ .

$$f'(x) <0$$ if $$0<x<3 \Rightarrow$$ $$f$$ is decreasing  in (1,3)

$$f'(x) >0$$ if $$x>3 \Rightarrow$$ $$f$$ is increasing  in $$(3, \infty)$$

$$\lim\limits_{x\to\infty} f(x) =2 \Rightarrow \,y=2$$ is horizontal asymptote.

$$f''(x)<0$$ in $$[0,2) \,\,\,\,{\text {&}}\,\,\,\, (5,\infty) \Rightarrow$$ Concave drown there.

$$f''(x) >0$$ in $$(2,5) \Rightarrow$$ Concave upward there

$$\therefore$$ Required option is $$'a'$$ .

Sketch the graph of a function $$f$$ which satisfy the following condition . $$f'(3)=0, \,\,f(3) =-2, \,\,f(0) =0$$ $$f'(x)<0$$ if  $$0<x<3$$  $$\text {&}$$  $$f'(x)>0$$  if $$x>3$$ $$f''(x) <0$$ if $$0\leq x<2 \,\,\,or \,\,\,x>5 \,\,\,\, {\text {&}} \,\,\,\, f''(x) >0 \,\,if\,\,\,2<x<5$$ $$\lim\limits_{x\to\infty} f(x) =2$$ $$f(-x) = f (x) \forall x$$

A B C D Option A is Correct

Vertical Asymptotes

The line $$x =a$$ is called vertical asymptote of $$y=f(x)$$ if any one or more of the following is true.

1. $$\lim\limits _{x\to a} f(x) = \infty$$
2. $$\lim\limits _{x\to a^-} f(x) = \infty$$
3. $$\lim\limits _{x\to a^+} f(x) = \infty$$
4. $$\lim\limits _{x\to a} f(x) =- \infty$$
5. $$\lim\limits _{x\to a^+} f(x) = -\infty$$
6. $$\lim\limits _{x\to a^-} f(x) =- \infty$$

e.g. $$f(x) =\dfrac{2x+3}{x-1}$$  has a vertical asymptote at $$x = 1$$ .  • Thus from the graph it is clear that horizontal asymptote is  $$y=2$$ and vertical asymptote is  $$x=1$$.
• Normally a function $$'f'$$ has vertical asymptote  where its  denominator is 0.

Find the horizontal and vertical asymptote of the curve  $$y= \dfrac{x^2+x-2}{2\,x^2-5\,x-7}$$

A Horizontal asymptote $$\to y= \dfrac{1}{2},$$ Vertical asymptote $$\to x=-1,x= \dfrac{7}{2}$$

B Horizontal asymptote $$\to y= 2,$$ Vertical asymptote $$\to x=2,x= \dfrac{3}{2}$$

C Horizontal asymptote $$\to y= -2,$$ Vertical asymptote $$\to x=1,x= \dfrac{5}{2}$$

D Horizontal asymptote $$\to y= 1,$$ Vertical asymptote $$\to x=-2,x= \dfrac{5}{4}$$

×

For vertical asymptote put Denominator =0

$$\therefore \,2x^2-5x-7=0$$

$$\Rightarrow 2x^2-7x+2x-7=0$$

$$\Rightarrow (2x-7)(x+1) =0$$

$$\Rightarrow x=-1,x=\dfrac{7}{2}$$

For horizontal asymptote

$$=\lim\limits_{x\to\infty} \,\dfrac{x^2+x-2}{2x^2-5x+7}$$

$$=\lim\limits_{x\to\infty} \,\,\dfrac{1+\dfrac{1}{x}-\dfrac{2}{x^2}}{2-\dfrac{5}{x}+\dfrac{7}{x^2}}$$          (Divide Numerator and Denominator by $$x^2$$)

$$= \dfrac{1+0-0}{2-0+0}=\dfrac{1}{2}$$

$$\therefore$$ Horizontal asymptote is $$y= \dfrac{1}{2}$$

$$=\lim\limits_{x\to-\infty} \,\dfrac{x^2+x-2}{2x^2-5x+7}$$

$$=\lim\limits_{x\to-\infty} \,\,\dfrac{1+\dfrac{1}{x}-\dfrac{2}{x^2}}{2-\dfrac{5}{x}+\dfrac{7}{x^2}}$$          (Divide Numerator and Denominator by $$x^2$$)

$$= \dfrac{1+0-0}{2-0+0}=\dfrac{1}{2}$$

$$\therefore$$ Horizontal asymptote is $$y= \dfrac{1}{2}$$

$$\therefore$$ Horizontal asymptote $$\to y=\dfrac{1}{2}$$

Vertical asymptote $$\to x=-1, x=\dfrac{7}{2}$$

Find the horizontal and vertical asymptote of the curve  $$y= \dfrac{x^2+x-2}{2\,x^2-5\,x-7}$$

A

Horizontal asymptote $$\to y= \dfrac{1}{2},$$ Vertical asymptote $$\to x=-1,x= \dfrac{7}{2}$$

.

B

Horizontal asymptote $$\to y= 2,$$ Vertical asymptote $$\to x=2,x= \dfrac{3}{2}$$

C

Horizontal asymptote $$\to y= -2,$$ Vertical asymptote $$\to x=1,x= \dfrac{5}{2}$$

D

Horizontal asymptote $$\to y= 1,$$ Vertical asymptote $$\to x=-2,x= \dfrac{5}{4}$$

Option A is Correct

Finding a Formula for a Function which Satisfies some Conditions about Asymptotes

Suppose we wish to find expression for function which has given horizontal and vertical asymptote then

1. Say  $$x = a$$  and $$x = b$$ are vertical asymptote we put the denominator of $$f(x)$$ as $$(x-a)(x-b) ×$$ (some other function).

2. Say $$y = c$$  is the horizontal asymptote then we should have $$\lim\limits_{x\to\infty} \,f(x) =c$$

Adjust the numerator of $$f(x)$$ so that above condition is satisfied.

Find a formula for function $$f$$  that has vertical asymptote $$x =2, \,\,x=4$$ and horizontal asymptote $$y=-1$$ .

A $$f(x) = \dfrac{-x^2+4x+3}{(x-2)(x-4)}$$

B $$f(x) = \dfrac{x^2+7\,x+6}{(x-1)(x-7)}$$

C $$f(x) = \dfrac{2\,x^2+x-3}{(x-1)(x-2)}$$

D $$f(x) = \dfrac{x^2+8\,x+3}{(x-4)(x+1)}$$

×

Vertical asymptote are $$x =2 \,\,\, {\text{&}}\,\,\,x=4$$

$$\Rightarrow$$ Denominator of the negative function = $$(x-2) (x-4)$$

$$\therefore\,\, f(x)$$  is of the form $$\dfrac{g(x)}{(x-2)(x-4)}$$

where $$g(x)$$ is some function .

Now $$\lim\limits_{x\to\infty} f(x) = -1 \Rightarrow g(x)$$ should be a quadratic .

expression with coefficient of $$x^2$$ as $$-1$$

$$\therefore g(x) = -x^2+4x+3$$        (4 and 3 are arbitrarily values)

$$\therefore f(x)= \dfrac{-x^2+4x+3}{(x-2)(x-4)}$$  or  $$f(x)= \dfrac{-x^2+\alpha x+\beta}{(x-2)(x-4)}$$

for any real $$\alpha$$ and  $$\beta$$ .

Find a formula for function $$f$$  that has vertical asymptote $$x =2, \,\,x=4$$ and horizontal asymptote $$y=-1$$ .

A

$$f(x) = \dfrac{-x^2+4x+3}{(x-2)(x-4)}$$

.

B

$$f(x) = \dfrac{x^2+7\,x+6}{(x-1)(x-7)}$$

C

$$f(x) = \dfrac{2\,x^2+x-3}{(x-1)(x-2)}$$

D

$$f(x) = \dfrac{x^2+8\,x+3}{(x-4)(x+1)}$$

Option A is Correct

Curve Sketching for Algebraic Functions

To  sketch the graph of any function we follow the steps given below . Some of the steps may not be relevant for some function .(e.g. Symmetry and asymptote)

1. Domain :

Find the domain of given function , this will give us the horizontal spam of the graph.

2. Intercepts :

$$y$$ intercept = $$f(0)$$ will tell the intersection points with $$x$$ axis  and  solving for $$x$$ by putting $$y=0$$ will give intersection with $$x$$ axis .(sometimes $$y=0$$  is difficult to solve on it this step in that case )

3. Symmetry :

(a) If $$f(-x) =f(x) \forall x$$  in the domain, then curve is symmetric about $$y$$ axis . Sketch the curve to right of axis and then take reflection in $$y$$ axis .

(b)  $$f(-x) =-f(x) \,\forall\, x$$  in the domain ,then curve is symmetric about origin. Sketch the graph to the right of origin and rotate  180º about origin.

4. Asymptote :

Find horizontal and vertical asymptotes.

5. Increase / Decrease interval :

Find the  interval of increase and decrease  of the function .

6. Local Maximum and Minimum Value :

Find the local maxima and minima points by calculating critical numbers  (i.e. $$f'(x) = 0$$ or $$f'(x)$$ does not  exist points). Then use first derivative test or second derivative test to find nature of extreme.

7. Concavity and Points of inflection :

Complete $$f''(x)$$  and use concavity test. Inflection points is the point where concavity changes.

8. Sketch the curve .

Use steps to sketch the curve  $$y= x^3 +3\,x^2 -9\,x+5$$ .

A B C D ×

Domains is R (all polynomials have domain R)

$$y$$ intercepts $$= f(0) =5\to$$ graph passes through (0,5)

$$x$$ intercepts cannot be found.

There is no symmetry as function is neither odd nor even.

There are no asymptote as $$\lim\limits _{x\to \infty} f(x) =\infty$$ and there is no Denominator .

For intervals of  increase or decrease

$$f'(x)>0 \Rightarrow 3x^2+6x-9 >0$$

$$\Rightarrow 3(x^2+2x-3) >0$$

$$\Rightarrow 3(x-1) (x+3) >0$$

$$\Rightarrow x \in (-\infty ,-3) \cup (1,\infty)$$

$$\therefore f$$ is increasing  in $$(-\infty ,-3) \cup (1,\infty)$$

$$f$$ is decreasing in $$(-3,1)$$

For local maxima and minima $$\to f'(x) = 0$$

$$\Rightarrow 3x^2 +6x-9 =0$$

$$\Rightarrow 3(x^2 +2x-3) =0$$

$$\Rightarrow 3(x-1) (x+3) = 0$$

$$\Rightarrow x= 1,-3$$

$$f''(x) = 6x+6 \Rightarrow f'' (1) = 12 >0$$

$$\therefore x=1$$ is a local minima

$$\Rightarrow f'' (-3) =-18 +6 =-12<0$$

$$\Rightarrow x=-3$$ is local maxima

$$f'' (x)>0 \Rightarrow x+1 >0$$

$$\Rightarrow x>-1$$

$$\therefore$$ Concave upwards in $$(-1,\infty)$$ and downward in $$(-\infty,-1)$$  . Also inflection points at $$x = -1$$.

Sketch the curve , Option 'a' is correct.

Use steps to sketch the curve  $$y= x^3 +3\,x^2 -9\,x+5$$ .

A B C D Option A is Correct

Sketch the graph of the curve $$y= \dfrac{x^2}{x^2-1}$$

A B C D ×

Domain of $$y$$ is $$R - \{x^2 -1 = 0 \,\,roots \}$$

$$\equiv R - \{-1,1\} \to$$ all real values except $$-1,1$$ .

$$y$$ intercept $$= f(0) = 0 \Rightarrow$$ graph passes through $$(0,0)$$

Put $$y=0 \Rightarrow \dfrac{x^2}{x^2-1}=0$$

$$\Rightarrow x=0 \Rightarrow$$graph passes through $$(0,0)$$

$$f(-x) = \dfrac{(-x)^2}{(-x)^2-1} = \dfrac{x^2}{x^2-1} \Rightarrow f(x)$$ is even

$$\Rightarrow$$ Symmetric about $$y$$ axis $$\to$$ we need the graph only to the right  $$y$$ axis .

For vertical  asymptotes $$\to$$ Denominator $$= x^2-1=0$$

$$\Rightarrow x=1,\,\, x=-1\to$$ Vertical asymptotes are $$x =\pm1$$

For horizontal asymptote

$$\lim\limits _{x\to\infty} \dfrac{x^2}{x^2-1} = \lim\limits_{x\to\infty} \dfrac{1}{1-\dfrac{1}{x^2}} = \dfrac{1}{1-0} =1$$

$$\therefore y=1$$ is a horizontal asymptotes.

For intervals of  increase or decrease

$$f'(x) >0 \Rightarrow \dfrac{(x^2-1)× 2x -x^2× 2x}{(x^2-1)^2}>0$$

$$\Rightarrow \dfrac{-2x}{(x^2-1)^2} >0\Rightarrow x<0$$

$$\Rightarrow f$$ is increasing in $$(-\infty,-1)\cup(-1,0)$$ , $$f$$ is decreasing in $$(0,1)\cup (1,\infty)$$

For maxima and minima $$\to$$ put $$f'(x)= 0$$

$$\Rightarrow -2x = 0 \Rightarrow x=0$$

$$f''(x)= \dfrac{d}{dx} \left(\dfrac{-2x}{(x^2-1)^2}\right) = \dfrac{(x^2-1)^2 × -2-(-2x)× 2(x^2-1)× 2x}{(x^2-1)^4}$$

$$\Rightarrow (x^2-1) × -2 \left[\dfrac{(x^2-1)-4x^2}{(x^2-1)^4}\right] = \dfrac{-2(-1-3x^2)}{(x^2-1)^3}$$

Now  $$f''(0) = \dfrac{-2× -1}{(-1)^3} =-2 <0$$     $$\therefore x=0$$ is a put of local maxima , there is no local minima.

$$f''(x)>0 \Rightarrow \dfrac{2(1+3x^2)}{(x^2-1)^3} >0$$

$$\Rightarrow x\in (-\infty,-1) \cup (1, \infty)$$

$$\therefore$$ concave upwards in $$(-\infty,-1) \cup(1,\infty)$$ and concave downward in $$(-1,1)$$ .

No point of inflection as $$f''(x) \neq 0$$ for any $$x$$ .

Sketch the curve , Option 'a' is correct.

Sketch the graph of the curve $$y= \dfrac{x^2}{x^2-1}$$

A B C D Option A is Correct