# McDonaldsArch

## Danielle M.

Here are two websites that may also help:

1. CHOOSE YOUR PARABOLA: For my parabola I choose to do half of the famous Mcdonald’s logo

2. DESCRIBE YOUR PARABOLA: The founder of this world famous restaurant, Mac McDonald and Dick wanted to start new franchise after the huge success of their first restaurant. Stanley Meston, the architect came up with a theme of red-and-white walk-up restaurant with a unique slanted roof. Dick didn’t like the design at first and said it was too simple. Thus, he suggested adding two huge yellow arches starting from the front of the restaurant to its back and to the both sides of the restaurant. They also incorporated another arch sporting “Speedee” logo to be identified as a sign for the restaurants. For many years, the arches only served as an architectural feature of the McDonalds.

In 1962, Dick thought to give a MacDonalds logo design more modernize form. The arches of McDonald’s when viewed from a particular angle looked like “M”. Thus, Dick and Mac McDonald used the arches in its new McDonalds logo. In point of fact, the early Golden arches were a diagonal line forming the roofline of McDonald’s restaurant, cutting the middle of “M”. It turned out to be a very huge success and the company stuck with the golden arches in the logo even as it began to get rid of them from its overall restaurants.

The latest McDonalds logo:

Nobody could have ever imagined what McDonalds logo graphic will turn out to be. The latest McDonald logo has incorporated the golden arches which seem to be the initial of McDonald that is “M”. It has become such famous that most of the people compare it
with “Swoosh” of the Nike and the unique Coca-Cola script. It is now an outright decipherable icon throughout the world.

3. GRAPH YOUR PARABOLA:

4. GIVE EQUATIONS FOR YOUR PARABOLA:
-Equation:To find my equation I choose to input 3 [(0.-11);(4,5); (8,-12)] coordinates into the formula f(x)/y = ax2 + bx + c. Making the equations:
- -11= 0a+0b+c ( here you can see that -11=c)
- 5= 4a+4b+c
- -12= 8a+8b+c
Since we can tell by the first equation that -11=c you simply go back and substitute in -11 for c, you should get the following new equations:
5=4a+4b= -11
-12= 8a+8b= -11
then you can multiply your top equation by -2 to cancel out a and solve for b. Then you go back in and substitute c and b and can simply solve for a, I got the equation
-(x-4)^2+5.
-(x-4)^2+5-
vertex form
-y= x^2-10x+29- standard form
- not factorable

1. VERTEX:The vertex of my parabola is (5,4). I found my vertex based off my equation. My vertex is a Maximum, I know this because this is the highest point that my graph reaches. The vertex of the graph signifies the middle of the parabola. There are two different formulas that you can use to find the axis of symmetry. One formula works when the parabola's equation is in vertex form and the other works when the parabola's equation is in standard form.
If your equation is in vertex form, then the axis of is
• x= h in the general vertex form equation y = (x-h)
2 + k
If your equation is in standard form, then the formula for the axis of symmetry is:
• x = -b/2a from the general standard form equation y = ax
2+bx + c

2. Y-INTERCEPT: The y-intercept of my graph is (0, -11). To find the y intercept without looking at the graph you would need to replace with 0 and solve. The y-intercept signifies where the line or in this case parabola crosses the y axis.
3. DISCRIMINANT: The discriminant of my parabola is 4^2-4(1)(5). The Discriminant (Δ) tells you the number of X-axis intercepts a polynomial function has.

If the discriminant:
= < 0, there are no x-axis intercepts.
= 0, there is one x-axis intercept.
= > 0, there are two x-axis intercepts.

4. X-INTERCEPT(S): My parabola has 2 x-intercepts and they are (0,1.7) and (0,6.3). x = 0 for the y-intercept(s), so replace x with o and solve. The x intercept tells you when the line or parabola crosses the x axis.

5. SOLUTIONS: The highest point that my graph will reach is (5,4)

7. Bonus:
-A parabola is A plane curve formed by the intersection of a right circular cone and a plane parallel to an element of the cone or by the locus of points equidistant from a fixed line and a fixed point not on the line. So Cones create parabola’s.