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?

1 Answer
Feb 23, 2017

Answer:

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