For my chapter 12 project I am describing the beam of a flashlight as a parabola. The flashlight as it is today was patented on august 26 1903 by Conrad Hubert. The vertex of my parabola is (0,-5), the x intercepts are -5 and 5, and is a maximum meaning that the a term is a positive number. The frame limits should be (-17,17)

Equations:
Slope Intercept, y=(1x-5)^2
To figure out the slope intercept form of my graph I first found out the slope and plugged it in as m in the equation y=mx+b, next I found my y intercept (-5) and plugged it in as b.

Standard Form, 0=x^2-y-25
To get my equation from slope intercept form to standard form I first simplified the equation to y=x^2-25, then I subtracted y from each side to get 0=x^2-y-25

Vertex form , y=x^2-5 or y=(x-0)^2-5
To get my equation into vertex form I simply plugged my graph into the equation y=(x-h)^2+k, the vertex of my graph is (0,-5) so I plugged 0 in for h and -5 in for k. Finally I simplified the equation and it came out to y=x^2-5 or y=(x-0)^2-5

## Flashlight Beam

## Parabola Project

## Adrian B.

Adrian Bezanis

Chapter 12 Project

For my chapter 12 project I am describing the beam of a flashlight as a parabola. The flashlight as it is today was patented on august 26 1903 by Conrad Hubert. The vertex of my parabola is (0,-5), the x intercepts are -5 and 5, and is a maximum meaning that the a term is a positive number. The frame limits should be (-17,17)

Equations:

Slope Intercept, y=(1x-5)^2

To figure out the slope intercept form of my graph I first found out the slope and plugged it in as m in the equation y=mx+b, next I found my y intercept (-5) and plugged it in as b.

Standard Form, 0=x^2-y-25

To get my equation from slope intercept form to standard form I first simplified the equation to y=x^2-25, then I subtracted y from each side to get 0=x^2-y-25

Vertex form , y=x^2-5 or y=(x-0)^2-5

To get my equation into vertex form I simply plugged my graph into the equation y=(x-h)^2+k, the vertex of my graph is (0,-5) so I plugged 0 in for h and -5 in for k. Finally I simplified the equation and it came out to y=x^2-5 or y=(x-0)^2-5