How to graph a parabola \[y = \dfrac{3}{2}{x^2} - 3x - \dfrac{5}{2}\]?

Hint:
Here, we will first convert the given equation into vertex form of the parabola by using the method of completing the square. Then we will compare the obtained equation with the general equation of parabola to find the vertex. Then we will find \[x\] and \[y\] intercepts and also an additional point by substituting different values of the variables in the vertex form of the given equation. Using this we will plot those points on a graph and join them together to find the required graph of the given parabola.

Complete step by step solution:
The given quadratic equation is \[y = \dfrac{3}{2}{x^2} - 3x - \dfrac{5}{2}\].
Now, we will use the method of completing the square.
\[\dfrac{3}{2}{\left( {x - 1} \right)^2} = \dfrac{3}{2}\left( {{x^2} - 2x + 1} \right) = \dfrac{3}{2}{x^2} - 3x + \dfrac{3}{2}\]
Hence, we can write the given quadratic equation as:
\[y = \dfrac{3}{2}{x^2} - 3x + \dfrac{3}{2} - \dfrac{3}{2} - \dfrac{5}{2}\]
\[ \Rightarrow y = \dfrac{3}{2}{\left( {x - 1} \right)^2} - \dfrac{3}{2} - \dfrac{5}{2}\]
Thus, solving further, we get,
\[ \Rightarrow y = \dfrac{3}{2}{\left( {x - 1} \right)^2} - 4\]………………………………………..\[\left( 1 \right)\]
This is in vertex form.
We know that the general equation of a parabola is \[y = a{\left( {x - h} \right)^2} + k\], where \[\left( {h,k} \right)\] is the vertex.
Now, in order to graph a parabola of this quadratic equation, first we will find the vertex.
Also, since, \[\left( {x - h} \right)\] is the axis of symmetry, thus, \[x = 1\] is the axis of symmetry in this question.
Comparing equation \[\left( 1 \right)\] with the given equation with the general equation, we have,
Vertex \[\left( {h,k} \right) = \left( {1, - 4} \right)\]
Now, substituting \[x = 0\] in the equation \[\left( 1 \right)\], we get,
\[y = \dfrac{3}{2}{\left( {0 - 1} \right)^2} - 4\]
\[ \Rightarrow y = \dfrac{3}{2} - 4\]
Subtracting the terms by taking LCM, we get
\[ \Rightarrow y = \dfrac{{3 - 8}}{2} = \dfrac{{ - 5}}{2}\]
Hence, when \[x = 0\], \[y = \dfrac{{ - 5}}{2}\]
Therefore, the \[y\]-intercept is \[\left( {0,\dfrac{{ - 5}}{2}} \right)\]
Similarly, substituting \[y = 0\] in the equation \[\left( 1 \right)\], we get,
\[0 = \dfrac{3}{2}{\left( {x - 1} \right)^2} - 4\]
Adding 4 on both sides, we get
\[ \Rightarrow 4 = \dfrac{3}{2}{\left( {x - 1} \right)^2}\]
Multiplying both sides by \[\dfrac{2}{3}\], we get
\[ \Rightarrow \dfrac{8}{3} = {\left( {x - 1} \right)^2}\]
Taking square root on both sides, we get
\[ \Rightarrow x - 1 = \pm \sqrt {\dfrac{8}{3}} \]
Adding 1 on both sides, we get
\[ \Rightarrow x = 1 \pm \dfrac{{2\sqrt 2 }}{{\sqrt 3 }} \cong 1 \pm 1.633\]
Hence, when \[y = 0\], \[x = \left( {1 + 1.633} \right),\left( {1 - 1.633} \right)\]
Therefore, the \[x\]-intercepts are \[\left( { - 0.633,0} \right)\] and \[\left( {2.633,0} \right)\]
Now, we will find an additional point also.
Thus, let \[x = 2\]
Hence, we get,
\[y = \dfrac{3}{2}{\left( {2 - 1} \right)^2} - 4\]
\[ \Rightarrow y = \dfrac{3}{2} - 4 = \dfrac{{3 - 8}}{2} = \dfrac{{ - 5}}{2}\]
Hence, additional point is \[\left( {2, - \dfrac{5}{2}} \right)\]
Now, we will draw the graph of parabola, such that,
Vertex \[=\left( {1, - 4} \right)\]
The \[x\]-intercepts are \[\left( { - 0.633,0} \right)\] and \[\left( {2.633,0} \right)\].
The \[y\]-intercept is \[\left( {0,\dfrac{{ - 5}}{2}} \right)\].
An additional point is \[\left( {2, - \dfrac{5}{2}} \right)\].
Hence, the required graph is:

Hence, this is the required answer.

Note:
A parabola is a curve having a focus and a directrix, such that each point on parabola is at equal distance from them. The axis of symmetry of a parabola is a line about which the parabola is symmetrical. When the parabola is vertical, the line of symmetry is vertical. When a quadratic function is graphed in the coordinate plane, the resulting parabola and corresponding axis of symmetry are vertical.