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

`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]

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

Create a data table of values as shown below:

In the following table,

you can see Column 1 (with x values) and

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

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:

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

`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.