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