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Welcome to our Math lesson on **The Graph of a Function**, this is the fifth lesson of our suite of math lessons covering the topic of **Injective, Surjective and Bijective Functions. Graphs of Functions**, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

In the previous tutorials, we have often dealt with the graphs of functions. It is a geometrical representation of the set of all points (ordered pairs) which - when substituted in the function's formula - make this function true.

In general, for every numerical function f: X → R, the graph is composed of an infinite set of real ordered pairs (x, y), where x ∊ R and y ∊ R. Every such ordered pair has in correspondence a single point in the coordinates system XOY, where the first number of the ordered pair corresponds to the x-coordinate (**abscissa**) of the graph while the second number corresponds to the y-coordinate (**ordinate**) of the graph in that point.

The graph below has five points A, B, C, D and E highlighted. Each of these points represents an ordered pair (x, y). Looking at closer, we identify the two coordinates (abscissa and ordinate) of each of these points as follows: A(-1, ** 5/2**), B(0, 0), C(2, -2), D(4, 0) and E(5,

If you have studied our previous tutorial, you will already be familiar with the method used to find the formula of a function shown in the graph. In the specific case, it is clear that this is a quadratic graph, so it has a general formula

f(x) = ax^{2} + bx + c

Since for x = 0 we have f(0) = 0, it results that the constant c is 0. Therefore, the general form of this function becomes

f(x) = ax^{2} + bx

We can factorise the above function as

f(x) = x(ax + b)

The two coefficients a and b are found by solving the equation f(x) = 0. This equation is true for x = 0 or for ax + b = 0. Substituting the x-coordinate of point D yields

a ∙ 4 + b = 0

Thus,

b = -4a

Therefore, the function becomes

f(x) = ax^{2} - 4ax

Finally, we can find the coefficient a by substituting the coordinates of another known point (for example C) in the function's formula. Thus, for x = 2 and y = -2 we obtain

-2 = a ∙ 2^{2} - 4a ∙ 2

-2 = 4a - 8a

-2 = -4a

a =*-2**/**-4*

a =*1**/**2*

-2 = 4a - 8a

-2 = -4a

a =

a =

Hence, the coefficient b is

b = 4a

= 4 ∙*1**/**2*

= 2

= 4 ∙

= 2

Therefore, the function shown in the graph is

f(x) = *1**/**2* x^{2} - 2x

Find the function shown in the graph below.

From the figure, it is clear that this graph belongs to a quadratic function (because the graph is a parabola) where the coefficient a is negative (because the arms of this parabola are directed downwards). The general formula of this function therefore is

f(x) = ax^{2} + bx + c

The parabola has two x-intercepts (otherwise known as roots, or zeroes): x^{1} - 1 and x^{x} = 3. For these two values the value of the function is zero, so we can write it as a quadratic equation with one variable

ax^{2} + bx + c = 0

Using the Vieta's formulas

x_{1} + x_{2} = *-b**/**a* and x_{1} ∙ x_{2} = *c**/**a*

we obtain

-1 + 3 = *-b**/**a*

2 =*-b**/**a*

b = -2a

2 =

b = -2a

and

-1 ∙ 3 = *c**/**a*

c = -3a

c = -3a

Therefore, the function shown in the graph becomes

f(x) = ax^{2} - 2ax - 3a

= a(x^{2} - 2x - 3)

= a(x

To find the unknown coefficient a, we substitute the coordinates of another known point in the above formula. We can choose for example the y-intercept (0, 3). In this way, we obtain

3 = a(0^{2} - 2 · 0 - 3)

3 = a ∙ (-3)

a =*3**/**-3*

a = -1

3 = a ∙ (-3)

a =

a = -1

Hence, we have

f(x) = (-1) ∙ x^{2} - 2 ∙ (-1) ∙ x - 3 ∙ (-1)

f(x) = -x^{2} + 2x + 3

f(x) = -x

You have reached the end of Math lesson **16.2.5 The Graph of a Function**. There are 7 lessons in this physics tutorial covering **Injective, Surjective and Bijective Functions. Graphs of Functions**, you can access all the lessons from this tutorial below.

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- Continuing learning functions - read our next math tutorial: Basic Functions

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