Question

What is the the vertex of `y = 2x^2-6x`?

Answer 1

The vertex is at `(1.5, -4.5)`

Explanation:

You could do this by the method of completing the square to find vertex form. But we can also factorise.

The vertex lies on the line of symmetry which is exactly half-way between the two `x`-intercepts. Find them by making `y=0`

`2x^2-6x =y`

`2x^2-6x =0`

`2x(x-3)=0`

`2x=0 " "rarrx =0`

`x-3=0" "rarrx=3`

The `x`-intercepts are at `0 and 3`

The midpoint is at `x=(0+3)/2 =3/2= 1 1/2`

Now use the value of `x` to find `y`

`y = 2(3/2)^2 -6(3/2)`

`y =4.5-9 = -4.5`

The vertex is at `(1.5, -4.5)`

Answer 2

The vertex occurs at `(3/2, -9/2)`

Explanation:

We have:

`y=2x^2-6x`

which is a quadratic expression, with a positive coefficient if `x^2` and so we have a `uu` shaped curve rather than a `nn` shape curve.

Method 1:

We can complete the square:

`y=2{x^2-3x}`
`\ \ =2{(x-3/2)^2-(3/2)^2}`
`\ \ =2{(x-3/2)^2-9/4}`
`\ \ =2(x-3/2)^2-9/2`

In this form we not that the first term `2(x-3/2)^2 ge 0`, and thus has a minimum value of `0` when

`2(x-3/2)^2 = 0` when `x=3/2`

Wit this value of `x` we have:

`y = -9/2`

Method 2:

We can find the roots of the equation and use the fact that the vertex occurs at the midpoint the roots (by symmetry of quadratics)

For the roots, we have:

`2x^2-6x = 0`

`:. 2x(x-3) = 0`
`:. x=0, x=3`

And so the midpoint (the `x`-coordinate of the vertex) is given by:

`x=(0+3)/2 = 3/2`, (as before).

And we find the `y`-coordinate by direct evaluation with `x=3/2`:

`y = 2(3/2)^2-6(3/2)`
`\ \ = 2 * 9/4 -6 * 3/2`
`\ \ = 18/4-18/2`
`\ \ = -18/4`
`\ \ = -9/2`, (as before)

We can verify these results graphically:

graph{y=2x^2-6x [-10, 10, -5, 5]}

Answer 3

vertex is at (1.5,-4.5)

Explanation:

`y=2x(x-3)`

So this is x intercept form we can easily find the x values when y is equal to zero.

We know that when we multiply if either product is zero the whole thing is zero.

So
`0=2x`
and
`0=x-3`

So we know that x can be either 0 or 3 when y is zero.

We know that a parabola is symmetrical so half way between these points we will find the x value of the vertex.

So this is `(3+0)/2=1.5`

So 1.5 is the x co-ordinate of the vertex so put in into the function to get the y co-ordinate

`f(1.5)=2(1.5)(1.5-3)=3(-1.5)=-4.5`

vertex is at (1.5,-4.5)