Chapter+10+Algebra

.Parabolas and how to graph them.

Parabolas can curve down or up and an easy way to tell which way it curves is to look at the leading coefficient. If the leading coefficient is negative, then the vertex is maximum, which means that it curves down. If the leading coefficient is positive, then the vertex is minimum, which means that the parabola curves upward.

The most simple way to graph a parabola is to make a T-chart. On the left side, you put in values of x and on the left side, you put in values of y. Then as you put values of x, you plug in those values into the equation to get the y-value. Make sure to use negative, positive, and 0 as x values because it might not start curving upward or downward until before the smallest point you made. You have to make sure that your line looks like it started curving when you decide to stop.

An easy way to graph a parabola is to first find the axis of symmetry using the formula X= b/2a. The axis of symmetry is the point in the x-axis where the vertex is. After finding the axis of symmetry, plug in the number you got for x into the equation you are given and you get a value for y. The x and y value make a coordinate for the graph (x,y). The vertex is that point. The vertex is the point where the parabola turns around or starts to go the other way.After finding the vertex, you find the y-intercept by making x in the equation equal to 0. you get another value of y and the next coordinate is (x,y) when x is 0. You graph the vertex and the y-intercept and then use the axis of symmetry to find a third point. To do that you will have to look at the axis of symmetry, the line that goes through the point that you got from x=b/2a, and look how far away the y-intercept is from the vertex and make the third point the same distance away from the vertex with the y value staying the same. Then you connect the dots and you should get a parabola. This however, doesn't work for all of the quadratic equations.

A way that always works is using the quadratic formula. You would plug in the values of a, b, and c into the formula and then solve for x.

The quadratic formula may always work but sometime it's better to use another way because it either is easier or it can be spotted easily. Another thing you could do is complete the square.

First, I put the loose number on the other side of the equation: 2x – 4//x// – 8 = 0 2x – 4//x// = 8 Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved Then I look at the coefficient of the //x// -term, which is –4 in this case. I take half of this number (including the sign), giving me –2. Then I square this value to get +4, and add this squared value to both sides of the equation: 2x – 4//x// + 4 = 8 + 4 //2x// – 4//x// + 4 = 12 This process creates a quadratic that is a perfect square, and factoring gives me: (//x// – 2)2 = 12 (I know it's a "minus two" inside the parentheses because half of –4 is –2 . If you note the sign when you're finding one-half of the coefficient, then you won't mess up the sign when you're converting to squared-binomial form.) Now I can square-root both sides of the equation, simplify, and solve: (//x// – 2)2 = 12
 * ** completing the square to solve //x//2 – 4//x// – 8 = 0. **

﻿x-2= ± sqrt(12)

Then you add 2 to both sides and you get x= 2 ±sqrt(12)

So your answers are 2+sqrt(12) and 2-sqrt(12), or 5.46 and -1.46

Another thing that you can do is use the discriminant to find how **many** solutions are in the quadratic equation. The discriminant is. If the discriminant is positive, then there is two solutions. If the discriminant is 0, then there is one solution, and if the discriminant is negative, then there is no solution.

Word Problems:

The height is defined in terms of the width, so I'll pick a variable for "width", and then create an expression for the height. Let " //w// " stand for the width of the picture. The height //h// is 4/3 the width, so   //h// = (4/3)//w//. Then the area is //A// = //hw// = [(4/3)//w//][//w//] = (4/3)//w//2 = 192. I need to solve this "area" equation for the value of the width, and then back-solve to find the value of the height. (4/3) //w//2 = 192 //w//2 = 144 //w// = ± 12 Since I can't have a negative width, I can ignore the " //w// = –12 " solution. Then the width must be 12 and the height is //h// = (4/3)(12) = 16. **The enlargement will be** **12** **inches by** **16** **inches.**
 * **A picture has a height that is** **4/3** **its width. It is to be enlarged to have an area of** **192** **square inches. What will be the dimensions of the enlargement?**

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 * **A garden measuring** **12** **meters by** **16** **meters is to have a pedestrian pathway installed all around it, increasing the total area to** **285** **square meters. What will be the width of the pathway?**
 * The first thing I need to do is draw a picture. Since I don't know how wide the path will be, I'll label the width as " //x// ".

Looking at my picture, I see that the total width will be  //x// + 12 + //x// = 12 + 2//x//, and the total length will be    //x// + 16 + //x// = 16 + 2//x//.

Then the new area is given by: ||  || (12 + 2//x//)(16 + 2//x//) = 285 ||

192 + 56//x// + 4//x//2 = 285 4//x//2 + 56//x// – 93 = 0 This quadratic is messy enough that I won't bother with trying to use factoring to solve; I'll just go straight to the quadratic formula: Obviously the negative value won't work in this context, so I'll ignore it. Checking the original exercise to verify what I'm being asked to find, I notice that I need to have units on my answer: **The width of the pathway will be** **1.5** **meters.**

**You have to make a square-bottomed, unlidded box with a height of three inches and a volume of approximately** **42** **cubic inches. You will be taking a piece of cardboard, cutting three-inch squares from each corner, scoring between the corners, and folding up the edges. What should be the dimensions of the cardboard, to the nearest quarter inch?**

** When dealing with geometric sorts of word problems, it is usually helpful to draw a picture. Since I'll be cutting equal-sized squares out of all of the corners, and since the box will have a square bottom, I know I'll be starting with a square piece of cardboard. **

I don't know how big the cardboard will be yet, so I'll label the sides as having length " //w// ".

Since I know I'll be cutting out three-by-three squares to get sides that are three inches high, I can mark that on my drawing.The dashed lines show where I'll be scoring the cardboard and folding up the sides.



Since I'll be losing three inches on either end of the cardboard when I fold up the sides, the final width of the bottom will be the original " //w// " inches, less three on the one side and another three on the other side. That is, the width of the bottom will be // w // – 3 – 3 = // w // – 6.

Then the volume of the box, from the drawing, is: ( // w // – 6)( // w // – 6)(3) = 42 ( // w // – 6)( // w // – 6) = 14 ( // w // – 6)2 = 14

This is the quadratic I need to solve. I can take the square root of either side, and then add the to the right-hand side:

...or I can multiply out the square and apply the quadratic formula:

Either way, I get two solutions which, when expressed in practical decimal terms, tell me that the width of the original cardboard is either about 2.26 inches or else about 9.74 inches. How do I know which solution value for the width is right? By checking each value in the original word problem. If the cardboard is only 2.26 inches wide, then how on earth would I be able to fold up three-inch-deep sides? But if the cardboard is 9.74 inches, then I can fold up three inches of cardboard on either side, and still be left with 3.74 inches in the middle. Checking: (3.74)(3.74)(3) = 41.9628 This isn't exactly 42, but, taking round-off error into account, it's close enough that I can trust that I have the correct value: **The cardboard should measure** **9.75** **inches on a side.**

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