Graph of a Qudratic Polynomial

                       Graph of a Quadratic polynomial

We know that any quadratic polynomial is given as $y=ax^2+bx+c$
It's the standard quadratic polynomial. The simplest form of the quadratic polynomial is $y=x^2$ which is obtained by substituting a=1, b=0 and c=0.
We can graph it using some coordinate points.
Let's assume $x=0$, we get $y=0^2=0$ , thus, coordinates of first point are  $(0, 0)$
Let's assume $x=1$, we get $y=1^2=1$, thus, coordinate of the second point are $(1, 1)$
Let's assume $x=-1$, we get $y=(-1)^2=1$, thus, coordinate of the third point are $(-1, 1)$
Let's assume $x=2$, we get $y=2^2=4$, thus, coordinate of the fourth point are $(2, 4)$
Let's assume $x=-2$, we get $y=(-2)^2=4$, thus, coordinate of the fourth point are $(-2, 4)$
Let's assume $x=3$, we get $y=3^2=9$, thus, coordinate of the fifth point are $(3, 9)$
Let's assume $x=-3$, we get $y=(-3)^2=4$, thus, coordinate of the fifth point are $(-3, 9)$

We can graph the parabola $y=x^2$ using the above points. We have the graph as shown below-

Now we can see that this graph is symmetric about y-axis.
Suppose, we have one point (x, y) on this graph then (-x, y) is also on the graph.
$Domain: (-\infty, \infty)$
$Range: [0, \infty)$
Axis of symmetry: $x=0$  or y-axis
Vertex: $(0, 0)$

Sometimes, graph of $y=x^2$ is called the graph of the parent quadratic polynomial.
And other quadratic polynomial can be expressed as the vertex form.
Vertex form is $y=a(x-h)^2+k$  where vertex is $(h, k)$ and axis of symmetry is $x=h$

Consider the polynomial $y=(x-2)^2+3$
Comparing this with the standard vertex form, we get $a=1, h=2\, and \,  k=3$
Thus, vertex of this parabola is $(2, 3).$
The graph of this parabola can be obtained by transforming the graph of $y=x^2$ as follows-
Transformations:
(i) 2 units right
(ii) 3 units up

Axis of symmetry is $x=2$
$Domain: (-\infty, \infty)$
$Range: [3, \infty)$
We get the parabola as shown in figure below: