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

##### 3 Answers

The vertex is at

#### 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

The

The midpoint is at

Now use the value of

The vertex is at

The vertex occurs at

#### Explanation:

We have:

# y=2x^2-6x #

which is a quadratic expression, with a positive coefficient if

**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 = 0# when#x=3/2#

Wit this value of

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

And we find the

# 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]}

vertex is at (1.5,-4.5)

#### Explanation:

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

and

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

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

vertex is at (1.5,-4.5)