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).

Jun 6, 2018

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]}