We have had a basic discussion on how to graph quadratic expressions, in the chapter on quadratic equations. Now, in terms of graph transformations, we will understand a stepbystep procedure to plot the graph of any quadratic function.
Consider the general quadratic function
\[f\left( x \right) = a{x^2} + bx + c\]
First, we rearrange it (by the method of completion of squares) to the following form:
\[f\left( x \right) = a{\left( {x + \frac{b}{{2a}}}\right)^2}  \frac{D}{{4a}}\]
The term D is the discriminant, given by \(D = {b^2}  4ac\). If you are unsure about this rearrangement, you are urged to refer to the chapter on quadratic equations.
Now, to plot the graph of f,we start by taking the graph of \({x^2}\) ,and applying a series of transformations on it:
Step1: \({x^2} \to a{x^2}\): This will imply a vertical scaling of the original parabola. If a is negative, the parabola will also flip its mouth from the positive to the negative side. The magnitude of the scaling depends upon the magnitude of a.
Step2: \(a{x^2} \to a{\left( {x + \frac{b}{{2a}}}\right)^2}\) : This is a horizontal shift of magnitude \(\left {\frac{b}{{2a}}} \right\) units. The direction of the shift will be decided by the sign of \(\frac{b}{{2a}}\). The new vertex of the parabola will be at \(\left( { \frac{b}{{2a}},0} \right)\). The following figure shows an example shift:
Step3: \(a{\left( {x + \frac{b}{{2a}}} \right)^2} \to a{\left( {x + \frac{b}{{2a}}} \right)^2}  \frac{D}{{4a}}\): This transformation is a vertical shift of magnitude \(\left {\frac{D}{{4a}}} \right\) units. The direction of the shift will be decided by the sign of \(\frac{D}{{4a}}\). The final vertex of the parabola will be at \(\left( {  \frac{b}{{2a}}, \frac{D}{{4a}}} \right)\). The following figure shows an example shift:
To summarize:

If a is positive, the parabola will open upward, else it will open downward.

The coordinates of the vertex of the parabola are \(\left({  \frac{b}{{2a}},  \frac{D}{{4a}}} \right)\).
Example 1: Plot the graph of \(f\left( x \right) = 1  2x 3{x^2}\).
Solution: We have:
\[\begin{array}{l}a =  3,\,\,\,b =  2,\,\,\,c = 1\\ \Rightarrow \,\,\,D = {b^2}  4ac =16\end{array}\]
The coordinates of the vertex are.
\[V \equiv \left( {  \frac{b}{{2a}}, \frac{D}{{4a}}} \right) = \left( {  \frac{1}{3},\frac{4}{3}} \right)\]
Note that the parabola will open downward (since a is negative), but the vertex has a positive ycoordinate.This means that the parabola crosses the xaxis. In other words, the quadratic function as real zeroes. Let us calculate these zeroes:
\[x = \frac{{  \left( {  2} \right) \pm \sqrt {16}}}{{  6}} =  1,\frac{1}{3}\]
The parabola will cross the xaxis at these x values.
Finally, using all this information,we plot the graph:
Example 2: Plot the graph of \(f\left( x \right) = 5x  4 {x^2}\).
Solution: We have:
\[\begin{array}{l}a =  1,\,\,\,b = 5,\,\,\,c =  4\\ \Rightarrow \,\,\,D = {b^2}  4ac =9\end{array}\]
Since a is negative, the parabola will open downward. The coordinates of the vertex are
\[V \equiv \left( {  \frac{b}{{2a}}, \frac{D}{{4a}}} \right) = \left( {\frac{5}{2},\frac{9}{4}} \right)\]
We can calculate the roots using the quadratic formula or through factorization; we use the latter approach in this case:
\[\begin{array}{l}f\left( x \right) = 5x  4  {x^2} =0\\ \Rightarrow \,\,\,{x^2}  5x + 4 =0\\ \Rightarrow \,\,\,\left( {x  1}\right)\left( {x  4} \right) = 0\\ \Rightarrow \,\,\,x = 1,4\end{array}\]
Using this information, we plot the graph of f: