Question

What is the vertex of `y= 4(x+2)^2-x^2-5x+3`?

Answer 1

The coordinate of the vertex is `(-11/6,107/12)`.

Explanation:

For the parabola given by the standard-form equation `y=ax^2+bx+c`, the `x`-coordinate of the parabola's vertex is at `x=-b/(2a)`.

So, to find the vertex's `x`-coordinate, we should first write the equation of this parabola in standard form. To do so, we have to expand `(x+2)^2`. Recall that `(x+2)^2=(x+2)(x+2)`, which can then be FOILed:

`y=4(x^2+2x+2x+4)-x^2-5x+3`

`color(white)y=4(x^2+4x+4)-x^2-5x+3`

Distribute the `4`:

`color(white)y=4x^2+16x+16-x^2-5x+3`

Group like terms:

`color(white)y=(4x^2-x^2)+(16x-5x)+(16+3)`

`color(white)y=3x^2+11x+19`

This is now in standard form, `y=ax^2+bx+c`. We see that `a=3,b=11`, and `c=19`.

So, the `x`-coordinate of the vertex is `x=-b/(2a)=-11/(2(3))=-11/6`.

To find the `y`-coordinate, plug `x=-11/6` into the parabola's equation.

`y=3(-11/6)^2+11(-11/6)+19`

`color(white)y=3(121/36)-121/6+19`

`color(white)y=121/12-121/6+19`

`color(white)y=121/12-242/12+228/12`

`color(white)y=107/12`

So, the coordinate of the vertex is `(-11/6,107/12)`.

graph{4(x+2)^2-x^2-5x+3 [-33.27, 31.68, -5.92, 26.56]}

Note that `(-11/6,107/12)approx(-1.83,8.92)`.