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.

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 \(x\).

- 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.

- 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.

- 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.

- 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.

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

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

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

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

- 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 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 \(a>0\))

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

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

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

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