Question

How do you identify if the equation \[y = {x^2} + 3x + 1\] is a parabola, circle, ellipse, or hyperbola? How do you graph it?

Answer 1

Hint: First we will try to see that which form of equation fits best for the given question, whether the equation of parabola, or equation of circle, or ellipse, or hyperbola. After that we will find the vertex form of the equation and solve it to get the points. Then we will plot the graph.

Complete step-by-step answer:
The given equation is:
 \[y = {x^2} + 3x + 1\]
We know that the equation of a parabola is in the form of:
 \[y = a{x^2} + bx + c\] .
According to the question, it is clear that the equation is the equation of parabola.
Now, we need to graph the equation. The best way to graph this equation is to convert the equation into the vertex form. The vertex form is:
 \[y = a{(x - h)^2} + k\] where (h, k) is vertex. We need to remember that a is at maxima when a>0, and a is at minima when a<0. When h=x, then it is the axis of symmetry of parabola.
The given equation is:
 \[y = {x^2} + 3x + 1\]
Now, we will modify the equation. We will add and subtract some terms, and also multiply some terms, and we get:
 \[ = {x^2} + 2 \times \dfrac{3}{2} \times x + {\left( {\dfrac{3}{2}} \right)^2} - {\left( {\dfrac{3}{2}} \right)^2} + 1\]
Here, we will rearrange some terms, and we get:
 \[ = {x^2} + {\left( {\dfrac{3}{2}} \right)^2} + 2 \times \dfrac{3}{2} \times x - {\left( {\dfrac{3}{2}} \right)^2} + 1\]
Here, we can see that the terms are in the form of the formula:
 \[{(a + b)^2} = {a^2} + {b^2} + 2ab\]
Here, a is ‘x’, b is \[\dfrac{3}{2}\] . So, with the help of this formula, we get:
 \[ = {\left( {{x^2} + \dfrac{3}{2}} \right)^2} - {\left( {\dfrac{3}{2}} \right)^2} + 1\]
Now, when we simplify it, we get:
 \[ = {\left( {x + \dfrac{3}{2}} \right)^2} - \dfrac{5}{4}\]
We can also write the answer in the form of:
 \[ = {\left( {x - \left( { - \dfrac{3}{2}} \right)} \right)^2} - \dfrac{5}{4}\]
Therefore, our vertex will be \[ - \dfrac{3}{2}, - \dfrac{5}{4}\] and axis of symmetry will be \[x + \dfrac{3}{2} = 0\]
Here, we will pick various values of ‘x’ around \[ - \dfrac{3}{2}\] , and then we will find ‘y’ from the given equation. We get points, and with the help of that we can plot the graph as following:

Note: There is another method to find whether the equation is a parabola, circle, ellipse or hyperbola. When we see the given equation, if \[{b^2} - 4ac = 0\] and a=0, c=0, then it is a parabola. If \[{b^2} - 4ac < 0\] and a=c, then it is a circle. If \[{b^2} - 4ac < 0\] , and \[a \ne c\] , then it is an ellipse. If \[{b^2} - 4ac > 0\] , then it is a hyperbola.