How to do you graph #y = 2x^2 + x - 3# by plotting points?

2 Answers
Jul 25, 2018

Vertex, #(-0.25,-3.125)#, axis of symmetry, #x= -0.25#
x intercepts, #(-1.5,0) and (1,0)#, y intercept, # (0 ,-3)#,
additional point, # (-2,3)#

Explanation:

#y= 2 x^2+x-3 or y= 2(x^2+0.5 x)-3 # or

#y= 2{x^2+0.5 x+(0.25)^2}-0.125-3 # or

#y= 2(x+0.25)^2-3.125 # ,this is vertex form of equation ,

#y=a(x-h)^2+k ; (h,k)# being vertex , here

#h=-0.25 ,k=-3.125,a=2 #, therefore vertex is at

# (-0.25, -3.125)#, since #a# is positive, parabola opens upward.

Axis of symmetry is #x= h or x = -0.25 ; # , y-intercept is found

by putting #x=0# in the equation #y=2 x^2_x-3 or y=-3#

y intercept is at # (0 ,-3)#

x-intercepts are found by putting #y=0# in the equation

# 2 x^2+x-3=0 or 2 x^2 -2 x +3 x -3= 0# or

#2 x( x-1) +3(x-1)=0 or (x-1) (2 x+3) =0 #

#:. x= 1 , x= -3/2 =-1.5# , x-intercepts are at

#(-1.5,0) and (1,0)#. Additional point:

#x=-2, y = 2 (-2)^2+ (-2)-3= 3 or (-2,3)#

graph{2x^2+x-3 [-10, 10, -5, 5]} [Ans]

Jul 25, 2018

#" "#
Please read the explanation.

Explanation:

#" "#
We have the quadratic equation #color(red)(y=f(x)=2x^2+x-3#

Create a data table of values as shown below:

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In the following table,

you can see Column 1 (with x values) and

Column 4 (with the corresponding y values) are shown:

enter image source here

Plot all the values from the table, to create the graph:

Obviously, from the table, for the value of #color(red)(y=0)#,

there are two corresponding #color(red)(x# values.

Hence, #color(blue)((-1.5,0) and (1.0,0))# are the x-intercepts..

Values in the shaded area are shown on the graph:

enter image source here

We will examine the Vertex and the Axis of Symmetry in the following graph:

enter image source here

#color(red)("Vertex :" (-0.25,-3.125)#

#color(blue)((0, -3)# is the y-intercept..

Axis of Symmetry is at :#color(blue)((x=-2.5)#

Hope you find the solution process useful.