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Difference between revisions of "Graph of a function"

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=== Solution ===
 
=== Solution ===
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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 determine whether the graph is above or below the <math>x</math>-axis 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.
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The quadratic equation can be written as <math>(2x+3)(x-1)</math> making the roots <math>x=-\frac{3}{2}</math> and <math>x=1</math>. Since the coefficient of the term with the highest power (in this case <math>x^2</math>) is <math>2>0</math>, the graph is above the <math>x</math>-axis for <math>(-\infty, -\frac{3}{2})</math> and <math>(1, +\infty)</math> and below the <math>x</math>-axis for <math>(-\frac{3}{2}, 1)</math>. This allows the graph to be drawn as a smooth curve curve through the zeros using this information as a guideline:
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Likewise, a sign analysis on the intervals <math>\left(-\frac 32, 1\right)</math> and <math>(1, \infty)</math> allows the graph to be drawn as a smooth curve curve through the zeros using this information as a guideline:
 
  
 
<center>[[Image:Parabola1.PNG]]</center>
 
<center>[[Image:Parabola1.PNG]]</center>
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=== Solution ===
 
=== Solution ===
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First, find the zeros of the function. Note that if <math>x=1</math> or <math>x=2</math>, <math>y=0</math>. After [[synthetic division]], the polynomial reduces to <math>y=(x-1)(x-2)(x^2+x-6)</math>.  Factor the quadratic gives <math>(x-1)(x-2)^2(x+3)</math>. So the roots are 1 and -3 and a double root at 2.  The final graph looks like:
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First, we need to find the zeros of the function. Notice that if <math>x=1</math> or <math>x=2</math>, <math>y=0</math>. Hence, the polynomial reduces to <math>y=(x-1)(x-2)(x^2+x-6)</math>.  Factoring the quadratic gives <math>(x-1)(x-2)^2(x+3)</math>. So the roots are <math>1</math> and <math>-3</math> and a double root at <math>2</math>.  The final graph looks like:
  
 
<center>[[Image:Quartic1.PNG]]</center>
 
<center>[[Image:Quartic1.PNG]]</center>
  
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== Sea also ==
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== See also ==
 
* [[Algebra]]
 
* [[Algebra]]

Latest revision as of 04:11, 8 June 2023

A graph is a visual representation of a function. If $f(x) = y$ then the point $(x,y)$ lies on the graph of $f$.

Graphing Points

A single point is the simplest thing to graph. The graph of $(2,5)$ would be a dot 2 units to the right of $y$-axis and 5 units above the $x$-axis.

Point(2,5).PNG

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 $2x + 3y = 24$.

Solution

To graph a line, it is necesasry to find two points $(x,y)$ that satisfy $2x + 3y = 24$. Letting $x=0$ gives $3y = 24\Leftrightarrow y = 8$. So $(0,8)$ is one point on the graph.

Find another point by letting $y=0$. Plugging this in and solving gives $x=12$. So $(12,0)$ is our other point.

Now plot these in the coordinate plane and draw a line through them:

Twopoints2.PNG

The arrowheads on the ends of the line segment indicate that the line goes on infinitely in both directions.

Graphing Polynomials

The first step in graphing a polynomial, $p(x)$, is to find the zeros of $p(x)=0$. 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 $x$-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 $y = 2x^{2} + x - 3$.

Solution

The quadratic equation can be written as $(2x+3)(x-1)$ making the roots $x=-\frac{3}{2}$ and $x=1$. Since the coefficient of the term with the highest power (in this case $x^2$) is $2>0$, the graph is above the $x$-axis for $(-\infty, -\frac{3}{2})$ and $(1, +\infty)$ and below the $x$-axis for $(-\frac{3}{2}, 1)$. This allows the graph to be drawn as a smooth curve curve through the zeros using this information as a guideline:

Parabola1.PNG

Problem

Graph $y = x^4 - 2x^3 -7x^2 +20x -12$.

Solution

First, we need to find the zeros of the function. Notice that if $x=1$ or $x=2$, $y=0$. Hence, the polynomial reduces to $y=(x-1)(x-2)(x^2+x-6)$. Factoring the quadratic gives $(x-1)(x-2)^2(x+3)$. So the roots are $1$ and $-3$ and a double root at $2$. The final graph looks like:

Quartic1.PNG

See also

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