# How do you identity if the equation y=x^2+3x+1 is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Feb 28, 2017

It is a parabola.

#### Explanation:

A equation of the form $y = a {x}^{2} + b x + c$ is a parabola, as such $y = {x}^{2} + 3 x + 1$ too is a parabola.

Another way is to look at general form of equation $A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$

If ${B}^{2} - 4 A C = 0$ and $A = 0$ or $C = 0$, it is a parabola

${B}^{2} - 4 A C < 0$ and $A = C$, it is a circle

${B}^{2} - 4 A C < 0$ and $A \ne C$, it is an ellipse

${B}^{2} - 4 A C > 0$, it is a hyperbola

Here $A = 1$, $B = 0$ and $C = 0$ and hence ${B}^{2} - 4 A C = 0$ and $C = 0$. Therefore, it is a parabola.

The easiest way to graph such an equation is to convert it into vertex form i.e. $y = a {\left(x - h\right)}^{2} + k$. In such a case $\left(h , k\right)$ is the vertex (if $a < 0$ it is at maxima and if $a > o$ it s at minima) and $x = h$ is axis of symmetry of parabola.

$y = {x}^{2} + 3 x + 1$

= ${x}^{2} + 2 \times \frac{3}{2} \times x + {\left(\frac{3}{2}\right)}^{2} - {\left(\frac{3}{2}\right)}^{2} + 1$

= ${\left(x + \frac{3}{2}\right)}^{2} - \frac{5}{4} = {\left(x - \left(- \frac{3}{2}\right)\right)}^{2} - \frac{5}{4}$

Hence, vertex is $\left(- \frac{3}{2} , - \frac{5}{4}\right)$ and axis of symmetry is $x + \frac{3}{2} = 0$.

We can then choose different values of $x$ around $- \frac{3}{2}$ and find $y$ from $y = {x}^{2} + 3 x + 1$ to get points

$\left(0 , 1\right) , \left(1 , 5\right) , \left(2 , 11\right) , \left(- 1 , - 1\right) , \left(- 2 , - 1\right) , \left(- 3 , 1\right) , \left(- 4 , 5\right) , \left(- 5 , 11\right)$

and graph appears as follows:

graph{x^2+3x+1 [-12.745, 7.255, -1.88, 8.12]}