Question

Can you help me identify the vertex and axis of symmetry y=6 (x+3)^2-4

Answer 1

EXPLANATION

Vertex:

The vertex of an up-down facing parabola of the form:

`y=ax^2+bx+c`

is:

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

Rewrite

`y=6(x+3)^2-4`

in the form y=ax^2 +bx + c

Expanding:

`y=6x^2+36x+50`

The parabola params are:

a=6, b=36, c=50

`x_v=-(b)/(2a)`
`x_v=-(36)/(2\cdot6)`

Simplify:

`x_v=-3`

Plug in x_v = -3to find the y_v value

`y_v=6(-3)^2+36(-3)+50`
`y_v=-4`

Therefore the parabola vertex is (-3,-4)

a=6 so the vertex is a minimum.

Axis:

Parabola standard equation

4p(y-k) = (x-h)^2 is the standard equation for an up-down facing parabola with vertex at (h,k), and a focal length p

Rewrite y=6(x+3)^2-4 in the standard form:

Add 4 to both sides

y + 4 = 6(x+3)^2 -4+4

Refine

y+4 = 6(x+3)^2

Divide both sides by 6

`(y+4)/(6)=(6(x+3)^2)/(6)`

Simplify

`(y)/(6)+(2)/(3)=(x+3)^2_{}_{}`

Factor 1/6

`(1)/(6)(y+((2)/(3))/((1)/(6)))=(x+3)^2`

Simplify:

`(1)/(6)(y+4)=(x+3)^2`

Factor 4:

`4\cdot((1)/(6))/(4)(y+4)=(x+3)^2`

Simplify:

`4\cdot(1)/(24)(y-(-4))=(x-(-3))^2`

Therefore parabola properties are:

`(h,k)=(-3,-4),\text{ p=1/24}`

Parabola is of the form 4p(y-k)=(x-h)^2 and is symmetric around the y-axis.

Axis of symmetry is a line parallel to the y-axis wich intersects the vertex:

x = -3