## Quadratic Equations In Real Life Essay

## Quadratic Functions Essay

635 WordsMay 21st, 20123 Pages

When you are graphing quadratics, it is the same as graphing linear equations but, quadratics have the curvy line, called a parabola. When you are graphing your points, it is best to graph three or more points. You are really going to need to point three or more points, because if there are less than three you will not have a correct graph, graphing more than three will insure that your graph will be correct. The biggest number that they say you have to graph will most likely not be able to be graphed because most of the graphs will not be big enough to graph that point. If you happen to somehow forget that the line has to be curved, having those extra points graphed will help remind you that the line will be curved. If you’re a value is*…show more content…*

Don't skip stuff, and you should do fine. You also need to know, that if you get in to the habit of forgetting the square root sign until the end of the problem then you will mess up on the test or work sheet, or whatever you are working on. You also have to remember that your equation HAS to be equal to 0 before finishing the problem. The point at the bottom where the curves come out, that it the vertex. When you are using quadratic function, you are usually looking for an objects trajectory….like how far you threw a football across the field, or in this case how far a rifle was thrown and how high it was thrown. Given three points in the plane that have different first coordinates and do not lie on a line, there is exactly one quadratic function f whose graph contains all three points. The graph contains three points and a parabola that goes through all three. The main thing you need to know is how to find it and how to graph it, well that is all you basically need to know how to

Show More

In 1972, you could buy a Mercury Comet for about $3,200. Cars can depreciate in value pretty quickly, but a 1972 Comet in pristine condition may be worth a lot of money to a collector today.

Let the value of one of these Comets be modeled by the quadratic function *v*(*t*) = 18.75*t*^{2} – 450*t* + 3,200, where *t* is the number of years since 1972. When is the value of the function equal to 0 (what is an *x*-intercept), what was the car’s lowest value, and what was its value in 2010?

**The car’s value never dropped to 0, the lowest value was $500, and the car was worth $13,175 in the year 2010.** In this model, the *y*-intercept represents the initial value. When *t* = 0, the function is *v*(0) = 3,200, which corresponds to the purchase price.

Find the *x*-intercepts by solving 18.75*t*^{2} –450*t* + 3,200 = 0. Using the quadratic formula (you could try factoring, but it’s a bit of a challenge and, as it turns out, the equation doesn’t factor), you get –37,500 under the radical in the formula. You can’t get a real-number solution, so the graph has no *x*-intercept. The value of the Comet doesn’t ever get down to 0.

Find the lowest value by determining the vertex. Using the formula,

This coordinate tells you that 12 years from the beginning (1984 — add 12 to 1972), the value of the Comet is at its lowest. Replace the *t*’s in the formula with 12, and you get *v*(12) = 18.75(12)^{2} – 450(12) + 3,200 = 500.

The Comet was worth $500 in 1984. To find the value of the car in 2010, you let *t* = 38, because the year 2010 is 38 years after 1972. The value of the car in 2010 is *v*(38) = 18.75(38)^{2} – 450(38) + 3,200 = $13,175.

## Leave a Comment

(0 Comments)