Question

How do you identity if the equation `2x^2+3x-4y+2=0` is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Answer 1

The given equation is a Parabola.

Explanation:

Given -

`2x^2+3x-4y+2=0`

The general for of the conic equation can be written as

`Ax^2+Cy^2+Dx+Ey+F=0`

In such case -

If `A xx C = 0`then it is a parabola.

In the given equation `y^2` is absent. Hence its coefficient is `C=0`

The coefficeint of `x^2` is `A=2`

In the given equation `A xx C = 0`. Hence the given equation is a Parabola.

To graph the equation, solve it for `y`

`4y=2x^2+3x+2`
`y=0.5x^2+0.75x+0.5`

Find its vertex

`x=(-b)/(2a)=(-0.75)/(2 xx 0.5)=-0.75`

Take a few points on either side of `x=-0.75`
Calculate the corresponding `y` values

Tabulate them

Plot the point on a graph sheet. connect them with a smooth curve.