|
|
| (2 intermediate revisions by one other user not shown) |
| Line 1: |
Line 1: |
| − | A '''graph''' is a visual representation of a [[function]]. If <math>f(x) = y</math> then the [[point]] <math>(x,y)</math> lies on the graph of <math>f</math>.
| + | The term '''graph''' has multiple meanings. It can refer to: |
| | | | |
| − | == Graphing Points ==
| + | * [[Graph of a function]] |
| − | A single point is the simplest thing to graph. The graph of <math>(2,5)</math> would be a dot 2 units to the right of <math>y</math>-axis and 5 units above the <math>x</math>-axis.
| + | * [[Graph (graph theory)]] |
| | | | |
| − | <center>[[Image:Point(2,5).PNG]]</center>
| + | {{disambig}} |
| − | | |
| − | == Graphing Lines ==
| |
| − | Given two distinct points on a line, one can construct the whole line. So one way to graph a line given its equation is to just find two points on it and to draw a straight line through them.
| |
| − | | |
| − | === Problem ===
| |
| − | Graph the line <math>2x + 3y = 24</math>.
| |
| − | | |
| − | === Solution ===
| |
| − | To graph a line, it is necesasry to find two points <math>(x,y)</math> that satisfy <math>2x + 3y = 24</math>. Letting <math>x=0</math> gives <math>3y = 24\Leftrightarrow y = 8</math>. So <math>(0,8)</math> is one point on the graph.
| |
| − | | |
| − | Find another point by letting <math>y=0</math>. Plugging this in and solving gives <math>x=12</math>. So <math>(12,0)</math> is our other point.
| |
| − | | |
| − | Now plot these in the coordinate plane and draw a line through them:
| |
| − | | |
| − | <center>[[Image:Twopoints2.PNG]]</center>
| |
| − | | |
| − | The arrowheads on the ends of the line segment indicate that the line goes on [[infinite]]ly in both directions.
| |
| − | | |
| − | == Graphing Polynomials ==
| |
| − | The first step in graphing a [[polynomial]], <math>p(x)</math>, is to find the zeros of <math>p(x)=0</math>. Then a smooth curve should be drawn through the zeros accounting for multiple roots and making sure the signs match up (i.e. the graph is above the <math>x</math>-axis when the polynomial is positive and below it when the polynomial is negative). This process is best understood through examples.
| |
| − | | |
| − | === Problem ===
| |
| − | Graph the parabola <math> y = 2x^{2} + x - 3 </math>.
| |
| − | | |
| − | === Solution ===
| |
| − | Luckily the quadratic factors as <math>(2x+3)(x-1)</math> making the roots <math>x=-\frac 32</math> and <math>x=1</math>. The quadratic can only switch signs as its zeros. So picking one point less than <math>-\frac 32</math> and plugging it in will tell us whether the graph is above or below the <math>x</math>-asis for all <math>x</math> on the interval <math>\left(-\infty, -\frac 32\right).</math> Since <math>f(-2)=3</math> is positive, the graph is above the <math>x</math>-axis.
| |
| − | | |
| − | Likewise, we do a sign analysis on the intervals <math>\left(-\frac 32, 1\right)</math> and <math>(1, \infty)</math>, draw a smooth curve curve through the zeros using this information as a guideline:
| |
| − | | |
| − | <center>[[Image:Parabola1.PNG]]</center>
| |
| − | | |
| − | === Problem ===
| |
| − | Graph <math>x^4 - 2x^3 -7x^2 +20x -12</math>.
| |
| − | | |
| − | === Solution ===
| |
| − | First, we find the zeros of the function. Plugging in numbers, we find <math>x=1</math> works. After [[synthetic division]] we have <math>(x-1)(x^3 - x^2 - 8x + 12)</math>. Plugging numbers into the cubic, we find that <math>x=2</math> works. Thus, we have <math>(x-1)(x-2)(x^2+x-6)</math>. Finally, we can factor the quadratic giving us <math>(x-1)(x-2)^2(x+3)</math>. So we have roots at 1 and -3 and a double root at 2. Our final graph looks like:
| |
| − | | |
| − | <center>[[Image:Quartic1.PNG]]</center>
| |
| − | | |
| − | == See also ==
| |
| − | * [[Algebra]]
| |
