Question

How do you graph `y=2x^2 -5x -3`?

Answer 1

You can easily do this by inserting values in x and plotting some points.

Explanation:

You can graph this easily by making a table of x and y values. But first, we must find the vertex `(h,k)`.

To solve for the vertex, we'll start with this formula (`h` is the value of the abscissa of the vertex.):
`h=-b/(2a)`
`h=-[(-5)]/[2(2)]`
`h=5/4`

Now to find `k` (the ordinate of the vertex), we will simply plug in `5/4` to `x`.

`y=2x^2-5x-3`
`y=2(5/4)^2-5(5/4)-3`
`y=2(25/16)-25/4-3`
`y=25/8-25/4-3`
`y=25/8-50/8-3`
`y=-25/8-3`
`y=-25/8-24/8`
`y=-49/8`

The vertex is `(5/4,-49/8)`.

After plotting the vertex, just plug in other values into x and graph them. Start with simple ones such as the y-intercept (set `x` to `0`). Remember that this is a quadratic equation, so your graph must be a parabola.

graph{2x^2-5x-3 [-14.24, 14.24, -7.12, 7.12]}

Answer 2

y= (2x+1) (x-3) graph{2x^2-5x-3 [-6.07, 7.977, -6.18, 0.847]}

Explanation:

Cross factorise the quadratic equation to get the (2`x`+1) (`x`-3) equation.

Then, put:

2`x`+1=0
so make `x` the subject of the formula
`x`=`-1/2`

And then,
`x`-3=0
so move -3 over
`x`=3

So `-1/2` and 3 are your `x`- intercepts.

To find the `y`- intercept you need to complete the square, making the quadratic into the form of `a(x-h)^2` + k, so

1. Factorise `2x^2` - 5`x` with the coefficient of `2x^2` which is 2 (because `x` has to have a coefficient of only 1 !)
2. Which makes it 2(`x^2` - 2.5) - 3
3. Then 2 [ `(x-1.25)^2` - 1.56 ) ] -3
   You get 1.25 from dividing 2.5 by 2. You always have to divide the number by 2. Also, you get 1.56 (rounded up to 3 s.f) from squaring 1.25. These are all rules!
4. Then expand the brackets, so
   2`(x-1.25)^2`-3.12-3
5. 2`(x-1.25)^2`- 6.12

So your `y`- intercept is 1.25. (Ignore the negative! Always change it to positive)

And your minimum point will be (1.25,-6.12).