# How to Graph Quadratic Functions(Parabolas)?

To draw the graph of a function in a Cartesian coordinate system, we need two perpendicular lines xOy (where O is the point where x and y intersect) called "coordinate axes" and a unit of measurement.

A point in this system has two coordinates.

M(x, y): M is the name of the point, x is the abscissa and is measured on Ox and y is the ordinate and is measured on Oy.

The two coordinates represent the distances from the point to the two axes.

If we consider a function f: A -> B (A - domain of definition, B - codomain), then a point found on the graph of the function is of the form P(x, f(x)).

__Example__

f:A -> B, f(x) = 3x - 1

If x = 2 => f(2) = 3×2 - 1 = 5 => P(2, 5) ∈ Gf(where Gf is the graph of the function).

## The Quadratic Function - Parabola

Standard form: *f(x) = ax ^{2} + bx + c*

Vertex form: $f(x)=(a+\frac{b}{2a})^2-\frac{\Delta}{4a}$

where *Δ = b ^{2} - 4ac*

If a > 0, the minimum value of *f(x)* will be $-\frac{\Delta}{4a}$ which is obtained if $x=-\frac{b}{2a}$.
The graph will be a **convex parabola** which vertex (the point where the parabola turns) is $V(-\frac{b}{2a};-\frac{\Delta}{4a})$.

If a < 0, the maximum value of *f(x)* will be $-\frac{\Delta}{4a}$
which is obtained if $x=-\frac{b}{2a}$.
The graph will be a **concave parabola** whose vertex is $V(-\frac{b}{2a};-\frac{\Delta}{4a})$.

The parabola is symmetric about the line that intersects $x=-\frac{b}{2a}$,
which is called the **"axis of symmetry"**.

That is the reason why, when we assign values to *x*, we
choose values symmetric about $-\frac{b}{2a}$.

When drawing a graph, the points of intersection with the coordinate axes are very important.

|. A point found on the *Ox* axis is of the form *P(x, 0)* because the distance from it to *Ox* is 0. If the point is found both on *Ox* and the graph of the function, it is also of the form *P(x, f(x)) ⇒ f(x) = 0*.

**Thus, to find the coordinates of the points of intersection with the Ox axis, we must solve the equation f(x)=0.**
We get to the equation

*a*.

^{2}+ bx + c = 0
The solutions of the equation depend on the sign of *Δ = b ^{2} - 4ac*.

We have the following situations:

**1) Δ < 0**

the equation has no solutions in *R* (the set of real numbers) the graph doesn't intersect *Ox*. The form of the graph will be:

or

**2) Δ = 0**

the equation has two equal solutions $x_1=x_2=-\frac{b}{2a}$

The graph is tangent to the *Ox* axis in the vertex of the parabola. The form of the graph is:

or

**3) Δ > 0**

the equation will have two different solutions.

$x_1=\frac{-b-\sqrt{\Delta}}{2a}$ and $x_2=\frac{-b+\sqrt{\Delta}}{2a}$

The graph of the function will intersect the *Ox* axis in points *M(x ^{1}* and

*Ox*. The form of the graph will be:

or

||. A point found on the *Oy* axis is of the form *R(0, y)* because the distance from it to *Oy* is *0*. If the point is found both on *Oy* and the graph of the function, it is also of the form *R(x, f(x)) ⇒ x = 0 ⇒ R(0, f(0))*.

In the case of the quadratic function,

*f(0) = a×0 ^{2} + b×0 + c ⇒ R(0, c).*

## The necessary steps to draw the graph of a quadratic function

*f: R → R
f(x) = ax ^{2} + bx + c*

1. We draw a table of variables in which we write some important values for *x*.

2. We find out the coordinates of the vertex $V(-\frac{b}{2a};-\frac{\Delta}{4a})$.

3. We also write 0 in the table and the symmetric of 0 about $-\frac{b}{2a}$.

or

4. We determine the point of intersection with the *Ox* axis by solving the equation *f(x)=0* and we write *x _{1}* and

*x*in the table.

_{2}Δ > 0 ⇒

Δ < 0 ⇒ there are no points of intersection. In this case we'll choose two convenient values symmetric about $-\frac{b}{2a}$

Δ = 0 ⇒ the graph is tangent to *Ox* right in its vertex. We'll choose once again two convenient values symmetric about $-\frac{b}{2a}$.
To better determine the form of the graph we can also choose other pairs of values for *x*, but they must also be symmetric about $-\frac{b}{2a}$.

5. We write all these values in the coordinate system and we draw the graph by connecting the points.

Example 1

f: R → R

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

a = 1, b = -2, c = -3

Δ = b^{2} - 4×a×c = (-2)^{2} - 4×1×(-3) = 16

$-\frac{b}{2a}=\frac{2}{2}=1$
⇒ V(1; -4)

1. $-\frac{\Delta}{4a}=-\frac{16}{4}=-4$

2. f(0) = -3

The symmetric of 0 about 1 is 2.

f(2) = -3

^{2}- 2x - 3 = 0

Δ = 16 > 0

$x_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{2-4}{2}=-1$

$x_1=\frac{2+4}{2}=3$

We have found the points:

A(-1; 0)

B(0; -3)

V(1; -4)

C(2; -3)

D(3; 0)

The graph will be:

Example 2

f: R → R

f(x) = -x^{2} - 2x + 8

a = -1, b = -2, c = 8

Δ = b^{2} - 4×a×c = (-2)^{2} - 4×(-1)×8 = 36

$-\frac{b}{2a}=\frac{2}{-2}=-1$
⇒ V(-1; 9)

1. $-\frac{\Delta}{4a}=-\frac{-36}{-4}=9$

2. f(0) = 8

f(-2) = 8 (the symmetric of 0 about -1 is -2)

3. f(x) = 0 ⇒ -x^{2} - 2x + 8 = 0

Δ = 36

x_{1} = 2 and x_{2} = -4

A(-4; 0)

B(-2; 8)

V(-1; 9)

C(0; 8)

D(2; 0)

Example 3

f: R → R

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

a = 1, b = -4, c = 4

Δ = b^{2} - 4×a×c = (-4)^{2} - 4×1×4 = 0

$-\frac{b}{2a}=\frac{4}{2}=2$
⇒ V(2; 0)

1. $-\frac{\Delta}{4a}=0$

2. f(0) = 4

f(4) = 4 (the symmetric of 0 about 2 is 4)

3. f(x) = 0 ⇒ x^{2} - 4x + 4 = 0

Δ = 0

x_{1} = x_{2} = $-\frac{b}{2a}$ = 2

A(-2; 9)

B(0; 4)

V(2; 0)

C(4; 4)

D(5; 9)

Example 4

f: R → R

f(x) = -x^{2} + 4x - 5

a = -1, b = 4, c = -5

Δ = b^{2} - 4×a×c = 4^{2} - 4×(-1)×(-5) = 16 - 20 = -4

$-\frac{b}{2a}=\frac{-4}{-2}=2$
⇒ V(2; -1)

1. $-\frac{\Delta}{4a}=-\frac{-4}{-4}=-1$

2. f(0) = -5

f(4) = -5 (the symmetric of 0 about 2 is 4)

3. f(x) = 0 ⇒ -x^{2} + 4x - 5 = 0,
Δ < 0

The equation doesn't have any solution.
We have to choose values symmetric about 2

A(-1; -10)

B(0; 5)

V(2; -1)

C(4; -5)

D(5; -10)

If the domain of definition is not R (the set of real numbers), but an interval, we erase the part of the graph corresponding to the values of *x* that aren’t found in the interval. It is necessary to write the endpoints in the table.

Example 5

f: [0; +∞) → R

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

a = 1, b = -2, c = -3

Δ = b^{2} - 4×a×c = (-2)^{2} - 4×1×(-3) = 16

$-\frac{b}{2a}=1$
⇒ V(1; -4)

1. $-\frac{\Delta}{4a}=-4$

2. f(0) = -3

f(2) = -3 (the symmetric of 0 about 1 is 2)

3. f(x) = 0 ⇒ x^{2} - 2x - 3 = 0,
Δ = 16

x_{1} = -1 ∉ [0; ∞)

x_{2} = 3

A(0; -3)

V(1; -4)

B(2; -3)

C(3; 0)

### How to create a parabola - animated