`y^2=16x` Graph the equation. Identify the focus, directrix, and axis of symmetry of the parabola.

Parabola is curve  graph such that each ordered pair `(x,y)` is equidistant distance to the fixed line (directrix) and fixed point (foci). We have "p" units  as distance of foci or directrix from the vertex. A parabola with vertex at the origin `(0,0)`  follow standard formula as:

a) `x^2 =4py` when parabola opens upward

b) `x^2 =-4py ` when parabola opens  downward

c) `y^2 =4px` when parabola opens to the right 

d)  `y^2...

Parabola is curve  graph such that each ordered pair `(x,y)` is equidistant distance to the fixed line (directrix) and fixed point (foci). We have "p" units  as distance of foci or directrix from the vertex. A parabola with vertex at the origin `(0,0)`  follow standard formula as:

a) `x^2 =4py` when parabola opens upward

b) `x^2 =-4py ` when parabola opens  downward

c) `y^2 =4px` when parabola opens to the right 

d)  `y^2 =-4px` when parabola opens left.

The given equation `y^2=16x ` resembles the standard formula `y^2=4px` .

 Thus, the parabola opens to the right and  we may solve for p using:

`4p =16`

`(4p)/4 = 16/4`

`p =4`

 When  parabola opens sideways, that  means the foci and vertex will have the same values of `x` . We follow the properties of the parabola that opens to right as:

vertex at point `(h,k)`

foci at point `(p,k)`

 directrix at` x= k-p`

axis of symmetry: `y =k`

endpoints of latus rectum: `(p,2p)` and` (p, -2p)`

Using vertex `(0,0)` , we have `h =0` and` k=0` .

Applying  `k=0` and `p=4` , we get the following properties:

a) foci at point `( 4, 0)`

b) axis of symmetry: `y=0` .

c) directrix at `x= 0-4 or x=-4.`

d) Endpoints of latus rectum: `(4,2*4)` and `(4, -2*4)` simplify to `(4,8)` and `(4.-8)` .

To graph the parabola, we connect the vertex with the endpoints of the latus rectum and extend it at both ends. Please see the attached file for the graph of  `y^2=16x.`