Question

What is the axis of symmetry and vertex for the graph `y=-¼x^2-2x-6`?

Answer 1

(1) : The Axis of symmetry is the line #x+4=0, and,

(2) : The Vertex is `(-4,-2)`.

Explanation:

The given eqn. is, `y=-1/4x^2-2x-6, i.e.`

`-4y=x^2+8x+24, or, -4y-24=x^2+8x`,

and completing the square of the R.H.S., we have,

`(-4y-24)+16=(x^2+8x)+16`,

`:. -4y-8=(x+4)^2`.

`:. -4(y+2)=(x+4)^2....................(ast)`.

Shifting the Origin to the point `(-4,-2),` suppose that,

`(x,y)` becomes `(X,Y).`

`:. x=X-4, y=Y-2, or, x+4=X, y+2=Y.`

Then, `(ast)` becomes, `X^2=-4Y..............(ast')`.

We know that, for `(ast'),` the Axis of Symmetry & the Vertex are,

the lines `X=0,` and `(0,0),` resp., in the `(X,Y)` System.

Returning back to the original `(x,y)` system,

(1) : The Axis of symmetry is the line #x+4=0, and,

(2) : The Vertex is `(-4,-2)`.

Answer 2

Axis of symmetry: `-4`

Vertex: `(-4,-2)`

Explanation:

Given:

`y=-1/4x^2-2x-6`, is a quadratic equation in standard form:

where:

`a=-1/4`, `b=-2`, and `c=-6`

Axis of Symmetry: the vertical line that divides the parabola into two equal halves, and the `x`-value of the vertex.

In standard form, the axis of symmetry `(x)` is:

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

`x=(-(-2))/(2*-1/4)`

Simplify.

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

Multiply by the reciprocal of `-2/4`.

`x=2xx-4/2`

Simplify.

`x=-8/2`

`x=-4`

Vertex: maximum or minimum point of a parabola.

Substitute `-4` into the equation and solve for `y`.

`y=-1/4(-4)^2-2(-4)-6`

Simplify.

`y=-1/4xx16+8-6`

`y=-16/4+8-6`

`y=-4+8-6`

`y=-2`

Vertex: `(-4,-2)` Since `a<0`, the vertex is the maximum point and the parabola opens downward.

graph{-1/4x^2-2x-6 [-12.71, 12.6, -10.23, 2.43]}