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

Dec 13, 2017

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

(2) : The Vertex is $\left(- 4 , - 2\right)$.

#### Explanation:

The given eqn. is, $y = - \frac{1}{4} {x}^{2} - 2 x - 6 , i . e .$

$- 4 y = {x}^{2} + 8 x + 24 , \mathmr{and} , - 4 y - 24 = {x}^{2} + 8 x$,

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

$\left(- 4 y - 24\right) + 16 = \left({x}^{2} + 8 x\right) + 16$,

$\therefore - 4 y - 8 = {\left(x + 4\right)}^{2}$.

$\therefore - 4 \left(y + 2\right) = {\left(x + 4\right)}^{2.} \ldots \ldots \ldots \ldots \ldots \ldots . \left(\ast\right)$.

Shifting the Origin to the point $\left(- 4 , - 2\right) ,$ suppose that,

$\left(x , y\right)$ becomes $\left(X , Y\right) .$

$\therefore x = X - 4 , y = Y - 2 , \mathmr{and} , x + 4 = X , y + 2 = Y .$

Then, $\left(\ast\right)$ becomes, ${X}^{2} = - 4 Y \ldots \ldots \ldots \ldots . . \left(\ast '\right)$.

We know that, for $\left(\ast '\right) ,$ the Axis of Symmetry & the Vertex are,

the lines $X = 0 ,$ and $\left(0 , 0\right) ,$ resp., in the $\left(X , Y\right)$ System.

Returning back to the original $\left(x , y\right)$ system,

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

(2) : The Vertex is $\left(- 4 , - 2\right)$.

Dec 13, 2017

Axis of symmetry: $- 4$

Vertex: $\left(- 4 , - 2\right)$

#### Explanation:

Given:

$y = - \frac{1}{4} {x}^{2} - 2 x - 6$, is a quadratic equation in standard form:

where:

$a = - \frac{1}{4}$, $b = - 2$, 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 $\left(x\right)$ is:

$x = \frac{- b}{2 a}$

$x = \frac{- \left(- 2\right)}{2 \cdot - \frac{1}{4}}$

Simplify.

$x = \frac{2}{- \frac{2}{4}}$

Multiply by the reciprocal of $- \frac{2}{4}$.

$x = 2 \times - \frac{4}{2}$

Simplify.

$x = - \frac{8}{2}$

$x = - 4$

Vertex: maximum or minimum point of a parabola.

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

$y = - \frac{1}{4} {\left(- 4\right)}^{2} - 2 \left(- 4\right) - 6$

Simplify.

$y = - \frac{1}{4} \times 16 + 8 - 6$

$y = - \frac{16}{4} + 8 - 6$

$y = - 4 + 8 - 6$

$y = - 2$

Vertex: $\left(- 4 , - 2\right)$ 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]}