How do you find the vertex?

#y=x(x-2)#

2 Answers
Jun 6, 2018

See below

Explanation:

I assume you meant x(x-2)=0. Expressions like the one you wrote don't have vertices because they can't be graphed on a two-dimensional graph.
First, expand the expression on the left using the distributive property: #x^2-2x=0#

Now, you can write the expression in standard form: #ax^2+bx+c#. This looks like #x^2-2x+0=0#, so #a=1#, #b=-2#, and #c=0#.

To find the axis of symmetry, on which the vertex lies, use the equation #x=-b/(2a)#. When you substitute in the numbers from the equation, it's #x=-b/(2a) = -(-2)/(2*1) = 2/2 = 1#

To find the y-coordinate of the vertex, substitute the x-coordinate into the equation: #y=x^2-2x = (1)^2-(2*1) = 1 - 2 = -1#

So the vertex of #x(x-2) = 0# is #(1, -1)#.

The vertex is #(0,-2)#.

Explanation:

Given:

#y=x(x-2)#

Expand the right-hand side.

#y=x^2-2x#

This is a quadratic equation in standard form:

#y=ax^2+bx+c#,

where:

#a=1#, #b=-2#, #c=0#

To find the vertex of a quadratic equation in standard form, use the formula for the axis of symmetry to find the x-coordinate. Then substitute the #x#-coordinate into the equation and solve for #y# to get the #y#-coordinate.

#x=(-b)/(2a)#

#x=(-(-2))/(2*1)#

#x=2/2#

#x=1#

Substitute #1# for #x# and solve for #y#.

#y=1^2-2(1)#

#y=1-2#

#y=-1#

The vertex is #(1,-1)#.

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