Informative line

### Graphical Representation Of Functions

Learn function formula with a graph symmetric about the y-axis is an even function. Practice with examples of even or odd functions symmetry and vertical line test, vertical line of symmetry.

# Ways of Representing a Function

There are 4 ways of representing a function

(1) Verbally (just describe it in words)

(2) Numerically (tables of values, one column contains x and the other values of y or f(x)

(3) Visually (by a graph)

(4) Algebraically (by an explicit formula)

Note- Most of the times the fourth representation will be the most convenient ways of representing a function.

• In the visual (graphical) representation ordered pair of values  $$(x, f(x))$$ are plotted on a graph paper and we join these points to get graph of $$f(x)$$.
• The value of  $$y(f(x))$$ at a particular point  $$(x)$$ is the height of graph of that particular  $$x$$.

#### Consider the graph of a function 'f' (in red). What is the value of f(3)?

A –1

B 0

C 4

D 2

×

Identify x = 3 on x-axis

Now note the graph height above (3, 0)

The height is 2, so f(3) = 2

Note that this function is called a constant function as it takes the constant value 2 for all values of x.

### Consider the graph of a function 'f' (in red). What is the value of f(3)?

A

–1

.

B

0

C

4

D

2

Option D is Correct

# Finding Value of Function from the Graph

• Given the graph of a function we can find the value of f(x) for all values of x.
• Also we can find the values of  x  for which expression f(x) is zero , positive or negative.

#### Consider the graph of a function f. The value of f(1) as observed from this graph is

A 0

B –10

C 6

D 3

×

Look for the point (1, 0) on x-axis

Note the height of the graph above this point

The height is 3 so f(1) = 3

### Consider the graph of a function f. The value of f(1) as observed from this graph is

A

0

.

B

–10

C

6

D

3

Option D is Correct

# Vertical Line Test

• A curve in the x-y plane is graph of a function only if no vertical line intersects it more than once. Vertical lines are $$x = a$$, so when it intersects the graph say at $$(a, b)$$ then there is one value of y i.e. $$b$$.
• If it intersect twice or more than twice then there are two values of y for one value of $$x$$, therefore not a function. Imagine all vertical lines, if at least one intersects twice or more conclude that it is not a function.

#### Which of the following graph represents  $$y$$ as a function of  $$x$$?

A

B

C

D

×

A curve in the $$x\text- y$$ plane is a graph of a function only if no vertical line intersects it more than once.

Since, the vertical line cuts the graph twice that means it gives two values of $$y$$ for one value of $$x$$. Therefore y is not a function of $$x$$.

Hence Option (A) is incorrect.

Since, any vertical line cuts the graph at a single point only means it gives one value of y at one value of $$x$$. Thus $$y$$ is a function of $$x$$.

Hence option (B) is correct.

Since, the vertical line cuts the graph at three points, it  means that it gives three values of  $$y$$ . Thus $$y$$ is not a function of $$x$$.

Hence option (C) is  incorrect.

Since, the vertical line cuts the graph at three points, it  means that it gives three values of  $$y$$ at a single value of  $$x$$. Thus $$y$$ is not a function of $$x$$.

Hence option (D) is incorrect.

### Which of the following graph represents  $$y$$ as a function of  $$x$$?

A
B
C
D

Option B is Correct

# Symmetry of Graphs

• The graph of all even functions are symmetric about y-axis, while the graph of odd functions are symmetric about the origin.
• Consider this, for even functions

f(–x) = f(x) for all x in domain.

• For every point (h, k) on the graph of such a function (–h, k) will also be a point .Since (h, k) and (–h, k) are mirror images of each other in the y-axis, the entire graph of even function will also be symmetric about y-axis.

• For odd functions → f(–x) = –f(x) for all x in the domain.
• For every point (h, k) on the graph of f(x) there will be a point (–h, –k) on the graph. Since (h, k) and (–h, –k) are reflections of each other in origin. (origin is the mid point of (h, k) and (h, –k)).
• Also note that graph of odd functions pass through origin because (f(–x) = –f(x) ⇒ f(0) = –f(0) ⇒ 2 f(0) = 0 ⇒ f(0) = 0.

#### The graph of four functions are given below, which one of them is even?

A

B

C

D

×

For even functions there should be symmetry about y-axis, only option 'a' satisfies this.

### The graph of four functions are given below, which one of them is even?

A
B
C
D

Option A is Correct

# Graphs of two or more Functions on the same Graph Paper

• If graphs of two or more functions are plotted on same x-y axis, we can analyze more things about the relation between function and values of x where they are equal.
• f(x) = g(x) at the points where the two graphs intersect.

#### Consider the graph of two functions f(x) and g(x). f(x) → red line, g(x) → green line For what value of x is f(x) = 0.

A 7

B –7

C –4

D 5

×

The point where the value of y or f(x) is 0 is the point where graph cuts x axis.

Since graph of f(x) intersects x-axis at x = –4, we say f(–4) = 0.

Thus the value of x at which f(x)=0 is -4.

### Consider the graph of two functions f(x) and g(x). f(x) → red line, g(x) → green line For what value of x is f(x) = 0.

A

7

.

B

–7

C

–4

D

5

Option C is Correct

#### If (2, –3) is a point on the graph of an odd function then which of the following points will also be on the graph?

A $$(5,\,7)$$

B $$(8,\,2)$$

C $$(-2,\,3)$$

D $$(4,\,1)$$

×

For every point $$(h,\,k)$$ on odd function graph, there will be a point $$(-h,\,-k)$$ on the same graph.So if $$(2,\,-3)$$ is on the graph, so will be $$(-2,\,3)$$.

### If (2, –3) is a point on the graph of an odd function then which of the following points will also be on the graph?

A

$$(5,\,7)$$

.

B

$$(8,\,2)$$

C

$$(-2,\,3)$$

D

$$(4,\,1)$$

Option C is Correct

# Completing the graph using symmetry

• If we know the graph of an even function which lies to right of y-axis, we can construct the graph to the left using symmetry and vice versa.
• Similarly for odd function symmetry about origin can be used.
• Symmetry about origin will mean that whenever $$(h,k)$$ is a point on the graph then $$(-h,-k)$$ will also be a point on the graph and line joining them will have midpoint as origin.
• If we know the graph  $$x \geq 0$$ we get the remaining graph by rotating this portion through $$180^{\circ}$$ about origin.

#### Given the incomplete graph of a certain even function 'f', choose which of the following is the complete graph of f.

A

B

C

D

×

Assuming y-axis as the mirror take reflections, we get the complete graph  as shown.

### Given the incomplete graph of a certain even function 'f', choose which of the following is the complete graph of f.

A
B
C
D

Option A is Correct

# Finding the values of x for which a given f(x) is positive, negative or zero

• Given a function  $$f(x)$$ , to find the value of  $$x$$ for which $$f(x)$$  is positive we need to solve the inequality  $$f(x)>0$$.
• Similarly the values of  $$x$$ for which  $$f(x)$$  is negative are given by  $$f(x)<0$$.
•  The equation $$f(x)=0$$ gives the values of $$x$$for which $$f(x)$$ vanishes or has value 0
• e.g. $$f(x)=\dfrac {x-2}{x^2+1}$$  is positive when $$\dfrac {x-2}{x^2+1}>0$$ or  $$x-2>0$$  or  $$x>2$$  or  $$x\in(2,\infty)$$

($$x^2+1$$ is a positive for all $$x$$)

($$\dfrac {a}{b}>0$$ if  $$b>0$$  and   $$a>0$$)

#### Find the values of $$x$$for which $$f(x)=\dfrac {2x-1}{x^2+5}$$ is positive.

A $$x\in(2,\infty)$$

B $$x\in(5,\infty)$$

C $$x\in \left (\dfrac {1}{2},\infty\right)$$

D $$x\in (-2,\,\infty)$$

×

$$f(x)$$ is positive when  $$f(x)>0$$  and  $$\dfrac {a}{b}>0\;$$

$$\Rightarrow\,a>0, \;b>0$$ or  $$a<0, \;b<0$$

$$\therefore \;f(x)>0\Rightarrow \dfrac {2x-1}{x^2+5}>0$$

$$\Rightarrow2x-1>0$$

($$x^2+5$$ is positive for all values of $$x$$)

$$\Rightarrow x>\dfrac {1}{2}$$ or $$x\in \left ( \dfrac {1}{2},\infty\right)$$

### Find the values of $$x$$for which $$f(x)=\dfrac {2x-1}{x^2+5}$$ is positive.

A

$$x\in(2,\infty)$$

.

B

$$x\in(5,\infty)$$

C

$$x\in \left (\dfrac {1}{2},\infty\right)$$

D

$$x\in (-2,\,\infty)$$

Option C is Correct