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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  \(x\).

Illustration Questions

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)?

image
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.

Illustration Questions

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

image
A

0

.

B

–10

C

6

D

3

Option D 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.

Illustration Questions

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.

image
A

7

.

B

–7

C

–4

D

5

Option C 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.

Illustration Questions

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.

image

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.

image

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.

image

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.

image

Which of the following graph represents  \(y\) as a function of  \(x\)?

A image
B image
C image
D image

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.

Illustration Questions

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 image
B image
C image
D image

Option A is Correct

Illustration Questions

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.

Illustration Questions

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.

image

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

image
A image
B image
C image
D image

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   \(a>0\))

Illustration Questions

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

Practice Now