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Graph of a function

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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