Question

What is the vertex of `y=4x^2-2x+(x+3)^2`?

Answer 1

The vertex is `(-2/5,41/5)` or `(-0.4,8.2)`.

Explanation:

Given:

`y=4x^2-2x+(x+3)^2`

First you need to get the equation into standard form.

Expand `(x+3)^2` using the FOIL method.
https://www.mathsisfun.com/definitions/foil-method.html

`y=4x^2-2x+x^2+6x+9`

Collect like terms.

`y=(4x^2+x^2)+(-2x+6x)+9`

Combine like terms.

`y=5x^2+4x+9` is a quadratic equation in standard form:

`ax^2+bx+c`,

where:

`a=5`, `b=4`, `c=9`

The vertex is the maximum or minimum point of a parabola. Since `a>0`, the vertex is the minimum point of this parabola, and the parabola opens upward.

The x-coordinate of the vertex is the same as the axis of symmetry for a quadratic equation in standard form. The formula is:

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

`x=(-4)/(2*5)`

`x=-4/10`

Simplify.

`x=-2/5` or `-0.4`

To calculate the y-coordinate of the vertex, substitute `-2/5` for `x` in the equation and solve for `y`.

`y=5(-2/5)^2+4(-2/5)+9`

`y=5(4/25)-8/5+9`

`y=20/25-8/5+9`

Simplify `20/25` to `4/5`.

`y=4/5-8/5+9`

Multiply `9` by `5/5` to get an equivalent fraction with `5` as the denominator.

`y=4/5-8/5+9xx5/5`

`y=4/5-8/5+45/5`

Simplify.

`y=41/5` or `8.2`

The vertex is `(-2/5,41/5)` or `(-0.4,8.2)`.

graph{y=5x^2+4x+9 [-11.72, 13.59, 5.72, 18.38]}