Coordinate Planes and Graphs

A coordinate plane turns geometry into numbers: every point gets a pair of numbers, and every equation in two variables gets a picture, its graph. On this page we build the coordinate plane, plot points and then draw graphs of equations and functions. At the end we move, reflect and stretch graphs without computing new tables of values.

On this page: Coordinate plane · Ordered pairs · Plotting points · Quadrants · Graph of an equation · Intercepts · Symmetry · Graph of a function · Translations · Reflections · Stretching · Common mistakes · Practice

The Rectangular Coordinate System

A rectangular coordinate system (also called a Cartesian coordinate system) is a pair of perpendicular number lines, called the coordinate axes, placed so that they cross at their zero points. The common point is the origin $O.$ Usually one axis is horizontal, with the positive numbers to the right, and the other is vertical, with the positive numbers upward. Both axes normally use the same unit of length.

The horizontal axis is usually labeled with the letter $x$ and the vertical one with $y.$ Then the axes are called the $x$-axis and the $y$-axis, and the plane is called the $xy$-plane. This is only a convention: any letters may be used. In applications the letters often tell what is measured, for example $t$ for time and $s$ for distance. The plane is named after the letters of its horizontal and vertical axes: with the axes $u$ and $v$ it is the $uv$-plane, with $t$ and $s$ the $ts$-plane, and so on.

Ordered Pairs and the Coordinates of a Point

An ordered pair of real numbers is two real numbers written in a fixed order, $(a, b).$ The order matters: $(4, 3)$ and $(3, 4)$ are different pairs. Two ordered pairs are equal exactly when their first numbers are equal and their second numbers are equal.

Every point $P$ of the coordinate plane corresponds to exactly one ordered pair. Draw two lines through $P\text{:}$ one perpendicular to the $x$-axis and one perpendicular to the $y$-axis. If the first line meets the $x$-axis at the number $a$ and the second one meets the $y$-axis at the number $b,$ then $a$ is the $x$-coordinate of $P$ and $b$ is its $y$-coordinate, and we write $P(a, b).$ The $x$-coordinate is also called the abscissa and the $y$-coordinate the ordinate. Conversely, every ordered pair $(a, b)$ gives exactly one point.

The $x$-coordinate always comes first: $P(a, b)$ means $a$ units along the $x$-axis and $b$ units along the $y$-axis.

Plotting Points

To plot the point $P(a, b)$ means to find its place in the coordinate plane. Start at the origin, move $|a|$ units to the right if ${a\gt0}$ or to the left if ${a\lt0},$ and then $|b|$ units up if ${b\gt0}$ or down if ${b\lt0}.$ For the point $P(4, 3)$ we go $4$ units to the right and $3$ units up. The point $Q(3, 4)$ is a different point.

The next figure shows the points $P(2, 5),$ $Q(-4, 3),$ $R(-5, -2)$ and $S(4, -3).$ The dashed lines show the coordinates of each point.

A point with $y$-coordinate $0$ lies on the $x$-axis, for example $(3, 0),$ and a point with $x$-coordinate $0$ lies on the $y$-axis, for example $(0, -2).$ The origin is the point $O(0, 0).$

Quadrants

The coordinate axes divide the plane into four regions called quadrants. They are numbered counterclockwise with Roman numerals, starting with the upper right one. In each quadrant the signs of the coordinates are always the same:

Quadrant	I	II	III	IV
$x$	$+$	$-$	$-$	$+$
$y$	$+$	$+$	$-$	$-$

For example, $P(2, 5)$ lies in quadrant I, $Q(-4, 3)$ in quadrant II, $R(-5, -2)$ in quadrant III and $S(4, -3)$ in quadrant IV. A point on an axis, such as $(3, 0)$ or $(0, -2),$ does not belong to any quadrant.

The Graph of an Equation

To check whether a point lies on a graph, substitute its coordinates into the equation. The point $(2, 4)$ lies on the graph of ${y=x^2},$ because ${4=2^2}.$ The point $(3, 6)$ does not, because ${3^2=9\ne6}.$

Sketching a graph point by point

The simplest way to sketch a graph is to compute a table of values, plot the points and join them with a smooth curve.

Example 1. Sketch the graph of ${y=x^2}.$

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y=x^2$	$9$	$4$	$1$	$0$	$1$	$4$	$9$

We plot the seven points $(-3, 9),$ $(-2, 4),$ $(-1, 1),$ $(0, 0),$ $(1, 1),$ $(2, 4),$ $(3, 9)$ and join them with a smooth curve. The curve is called a parabola. (So that the point $(3, 9)$ fits into the figure, the unit on the $y$-axis is shorter than the unit on the $x$-axis.)

A few points only suggest the shape of a graph: we do not know for sure what happens between them. More points give a better picture, and calculus gives methods to find the exact shape, for example where a graph rises or falls and how it bends.

Example 2. Sketch the graph of ${y=\frac{1}{x}}.$

Division by zero is not defined, so ${x\ne0}\text{:}$ there is no point of the graph on the $y$-axis. We take positive and negative values of $x,$ including values close to $0.$

$x$	$\frac{1}{3}$	$\frac{1}{2}$	$1$	$2$	$3$
$y=\frac{1}{x}$	$3$	$2$	$1$	$\frac{1}{2}$	$\frac{1}{3}$
$x$	$-\frac{1}{3}$	$-\frac{1}{2}$	$-1$	$-2$	$-3$
$y=\frac{1}{x}$	$-3$	$-2$	$-1$	$-\frac{1}{2}$	$-\frac{1}{3}$

The points with ${x\gt0}$ lie in quadrant I and the points with ${x\lt0}$ in quadrant III. The graph has two separate pieces (branches), and we must not join them across the $y$-axis. When $x$ comes close to $0,$ the value $\frac{1}{x}$ becomes very large in absolute value, and when $|x|$ grows, $\frac{1}{x}$ comes close to $0.$ So the graph comes closer and closer to the axes but never touches them. This curve is called a hyperbola.

Intercepts

The points where a graph meets the coordinate axes are especially useful when we sketch it. An $x$-intercept of a graph is the $x$-coordinate of a point where the graph meets the $x$-axis, and a $y$-intercept is the $y$-coordinate of a point where the graph meets the $y$-axis.

Example 3. Find all intercepts of the graphs of a) ${3x+2y=6};$ b) ${x=y^2-2y};$ c) ${y=\frac{1}{x}}.$

a) The $x$-intercept: we set ${y=0}$ and get ${3x=6},$ so ${x=2}.$ The $y$-intercept: we set ${x=0}$ and get ${2y=6},$ so ${y=3}.$ The graph is the line through the points $(2, 0)$ and $(0, 3).$

b) The $x$-intercept: for ${y=0}$ we get ${x=0^2-2\cdot0=0},$ so the $x$-intercept is $0.$ The $y$-intercepts: for ${x=0}$ we get ${y^2-2y=0},$ that is, ${y(y-2)=0},$ so ${y=0}$ or ${y=2}.$ The graph meets the axes at $(0, 0)$ and $(0, 2).$ It is a parabola that opens to the right, with its vertex at $(-1, 1)\text{:}$ we can write ${x=(y-1)^2-1},$ so $x$ is smallest, ${x=-1},$ when ${y=1}.$

c) The $x$-intercepts: for ${y=0}$ we get ${\frac{1}{x}=0}.$ A fraction with numerator $1$ is never $0,$ so there is no solution and no $x$-intercept. The $y$-intercepts: ${x=0}$ is not allowed, because $\frac{1}{0}$ is undefined, so there is no $y$-intercept either. The graph of ${y=\frac{1}{x}}$ does not meet the axes, as we saw in Example 2.

Symmetry

In the figure below, the points $(x, y),$ $(-x, y),$ $(-x, -y)$ and $(x, -y)$ are the corners of a rectangle whose center is the origin. The point $(x, -y)$ is the reflection of $(x, y)$ about the $x$-axis, the point $(-x, y)$ is its reflection about the $y$-axis, and $(-x, -y)$ is symmetric to $(x, y)$ about the origin.

A symmetric graph is easier to draw: we sketch one half and get the other half by reflection. The definitions give simple tests. A graph is symmetric about the $x$-axis if replacing $y$ by $-y$ gives an equivalent equation, symmetric about the $y$-axis if replacing $x$ by $-x$ gives an equivalent equation, and symmetric about the origin if replacing both $x$ by $-x$ and $y$ by $-y$ gives an equivalent equation.

Example 4. Test the graphs of ${y=x^2},$ ${y=\frac{1}{x}}$ and ${x=y^2-2y}$ for symmetry.

For ${y=x^2}\text{:}$ replacing $x$ by $-x$ gives ${y=(-x)^2=x^2},$ the same equation. Replacing $y$ by $-y$ gives ${-y=x^2},$ that is, ${y=-x^2},$ which is a different equation: the point $(1, 1)$ satisfies the first one but not the second. Replacing both $x$ and $y$ gives ${-y=(-x)^2},$ that is, again ${y=-x^2},$ so there is no symmetry about the origin either.

For ${y=\frac{1}{x}}\text{:}$ replacing both $x$ and $y$ gives ${-y=\frac{1}{-x}},$ that is, ${y=\frac{1}{x}}$ again. Replacing only $x$ (or only $y$) gives ${y=-\frac{1}{x}},$ a different equation: the point $(1, 1)$ satisfies ${y=\frac{1}{x}}$ but not ${y=-\frac{1}{x}}.$

For ${x=y^2-2y}\text{:}$ the point $(0, 2)$ is on the graph, but $(0, -2)$ is not, because ${(-2)^2-2\cdot(-2)}={8}\ne{0}.$ The point $(-1, 1)$ is on the graph, but $(1, 1)$ is not, because ${1^2-2\cdot1=-1\ne1}.$ So the graph is not symmetric about the $x$-axis and not about the $y$-axis. It is not symmetric about the origin either: the reflection of $(0, 2)$ about the origin is also $(0, -2),$ which is not on the graph. (It is symmetric about the horizontal line ${y=1}$ through its vertex.)

Graph	Symmetric about
	the $x$-axis	the $y$-axis	the origin
${y=x^2}$	no	yes	no
${y=\frac{1}{x}}$	no	no	yes
${x=y^2-2y}$	no	no	no

The Graph of a Function

So the point $(a, b)$ lies on the graph of $f$ exactly when $a$ is in the domain of $f$ and ${b=f(a)}.$ The value $f(a)$ is the height of the graph above the point $a$ of the $x$-axis; if ${f(a)\lt0},$ the graph is $|f(a)|$ units below the $x$-axis there.

Example 5. Sketch the graph of ${f(x)=x+2}.$

We draw the graph of ${y=x+2}.$ For ${x=-2},$ $0$ and $2$ we get ${y=0},$ $2$ and $4.$ The points $(-2, 0),$ $(0, 2)$ and $(2, 4)$ lie on a line. This is no accident: the graph of ${f(x)=mx+b}$ is always a line with slope $m$ and $y$-intercept $b$ (see Line and Definition of Slope), so two points are enough to draw it.

Example 6. Sketch the graph of ${f(x)=|x|}.$

By the definition of the absolute value,

$$|x|=\begin{cases}x, & \text{if } x\ge0,\\ -x, & \text{if } x\lt0.\end{cases}$$

The absolute value of a non-negative number is the number itself, and the absolute value of a negative number is its opposite. So for ${x\ge0}$ the graph is the part of the line ${y=x}$ in quadrant I, together with the origin, and for ${x\lt0}$ it is the part of the line ${y=-x}$ in quadrant II. Putting the two halves together, we get a V with its corner at the origin. The graph is symmetric about the $y$-axis, because ${|{-x}|=|x|}.$

Example 7. Sketch the graph of ${t(x)=\frac{x^2-4}{x-2}}.$

The function is not defined at ${x=2},$ where the denominator is $0.$ For every other $x$ we can factor the numerator and cancel:

${t(x)}={\frac{x^2-4}{x-2}}={\frac{(x-2)(x+2)}{x-2}}={x+2}$   for ${x\ne2}.$

So the graph of $t$ is the line ${y=x+2}$ without the point $(2, 4).$ In a sketch, the missing point is shown as an open circle (a "hole").

Example 8. Sketch the graph of

$$g(x)=\begin{cases}1, & \text{if } x\le2,\\ x+2, & \text{if } x\gt2.\end{cases}$$

For ${x\le2}$ the graph is part of the horizontal line ${y=1}.$ It includes the point $(2, 1),$ because ${g(2)=1}\text{:}$ we draw it as a filled dot. For ${x\gt2}$ the graph is part of the line ${y=x+2}.$ It starts at $(2, 4),$ but this point is not on the graph, because for ${x=2}$ the first formula is used. We draw it as an open circle.

The vertical line test

A function gives only one value $f(a)$ for each $a,$ so the vertical line ${x=a}$ (the line of all points with $x$-coordinate $a$) can meet the graph of a function at most once. If this line met the graph of $f$ at two points $(a, b)$ and $(a, c)$ with ${b\ne c},$ then $f(a)$ would have to be both $b$ and $c,$ which is impossible.

Vertical line test. A curve in the $xy$-plane is the graph of a function ${y=f(x)}$ if and only if no vertical line meets it at more than one point.

For example, the curve ${x=y^2-2y}$ from Example 3 is not the graph of any function ${y=f(x)}\text{:}$ the vertical line ${x=3}$ meets it at $(3, 3)$ and $(3, -1),$ because ${3^2-2\cdot3=3}$ and ${(-1)^2-2\cdot(-1)=3}.$

Graphing Functions by Translations

Suppose we already know the graph of ${y=f(x)},$ and $c$ is a positive number. Then we can get the graphs of ${y=f(x)+c},$ ${y=f(x)-c},$ ${y=f(x+c)}$ and ${y=f(x-c)}$ without new tables of values: each of them is the graph of $f$ moved (translated) by $c$ units.

Equation	The graph of ${y=f(x)}$ is translated
${y=f(x)+c}$	up by $c$ units
${y=f(x)-c}$	down by $c$ units
${y=f(x+c)}$	left by $c$ units
${y=f(x-c)}$	right by $c$ units

Vertical translations: ${y=x^2},$ ${\textcolor{#2b7a2b}{y=x^2+2}}$ and ${\textcolor{#c0392b}{y=x^2-2}}.$ Horizontal translations: ${y=x^2},$ ${\textcolor{#2b7a2b}{y=(x+2)^2}}$ and ${\textcolor{#c0392b}{y=(x-2)^2}}.$

The vertical translations are easy to see: in ${y=f(x)+c}$ every $y$-coordinate is $c$ larger. The horizontal ones often look backwards. The function ${g(x)=f(x-c)}$ takes at the point ${x+c}$ the value ${g(x+c)}={f\big((x+c)-c\big)}={f(x)},$ the same value that $f$ takes at $x.$ So every point $(x, f(x))$ of the old graph moves to $(x+c, f(x)),$ which is $c$ units to the right. In the same way ${y=f(x+c)}$ moves the graph to the left.

Inside the parentheses the signs work the opposite way: ${y=(x-2)^2}$ is the parabola ${y=x^2}$ moved $2$ units to the right, and ${y=(x+2)^2}$ is moved $2$ units to the left.

Example 9. Sketch the graph of ${y=|x-3|+2}.$

We start with the graph of ${y=|x|}$ from Example 6. Translating it $3$ units to the right gives the graph of ${y=|x-3|}.$ Translating that graph $2$ units up gives the graph of ${y=|x-3|+2}.$ The corner moves from $(0, 0)$ to $(3, 0)$ and then to $(3, 2).$

Example 10. Sketch the graph of ${y=x^2-4x+5}.$

We complete the square. The coefficient of $x$ is $-4,$ half of it is $-2$ and ${(-2)^2=4},$ so we write $5$ as ${4+1}\text{:}$

${y}={x^2-4x+5}={(x^2-4x+4)+1}={{(x-2)^2+1}.}$

In this form we see that the graph is the graph of ${y=x^2}$ translated $2$ units to the right, because of the ${x-2},$ and $1$ unit up, because of the ${+1}.$ The vertex of the parabola is $(2, 1).$ For ${x=0}$ we get ${y=5},$ so the parabola crosses the $y$-axis at $(0, 5).$

Reflections

Indeed, if the point $(a, b)$ is on the graph of ${y=f(x)},$ that is, ${b=f(a)},$ then the point $(-a, b)$ satisfies ${y=f(-x)},$ and the point $(a, -b)$ satisfies ${y=-f(x)}.$ For example, the point $(4, 2)$ is on the graph of ${y=\sqrt{x}},$ the point $(-4, 2)$ is on the graph of ${y=\sqrt{-x}},$ and the point $(4, -2)$ is on the graph of ${y=-\sqrt{x}}.$

Example 11. Sketch the graph of ${y=\sqrt[3]{2-x}}.$

The graph can be obtained by a reflection and a translation:

1. Draw the graph of ${y=\sqrt[3]{x}}.$
2. Reflect it about the $y$-axis to get the graph of ${y=\sqrt[3]{-x}}.$
3. Translate this graph $2$ units to the right, that is, replace $x$ by ${x-2}.$ We get the graph of ${y=\sqrt[3]{-(x-2)}},$ which is the same as ${y=\sqrt[3]{2-x}}.$

Gray: ${y=\sqrt[3]{x}};$ orange: ${\textcolor{#d35400}{y=\sqrt[3]{-x}}};$ blue: ${\textcolor{#1b64a8}{y=\sqrt[3]{2-x}}}.$

A few points confirm the sketch: for ${x=-6},$ $1,$ $2,$ $3$ and $10$ the values of $\sqrt[3]{2-x}$ are $2,$ $1,$ $0,$ $-1$ and $-2.$

Do not read the ${+2}$ in ${\sqrt[3]{-x+2}}$ as a translation of ${y=\sqrt[3]{-x}}$ to the left. First factor out the minus sign: ${-x+2=-(x-2)}.$ So the graph of ${y=\sqrt[3]{-x}}$ is translated $2$ units to the right.

Stretching and Compressing

The points on the $x$-axis do not move, because ${c\cdot0=0}.$ For example, above ${x=1}$ the parabola ${y=x^2}$ has height $1$ and the parabola ${y=2x^2}$ has height $2.$ Above ${x=2}$ the parabola ${y=x^2}$ has height $4$ and the parabola ${y=\frac{1}{2}x^2}$ has height $2.$

${y=x^2},$ ${\textcolor{#c0392b}{y=2x^2}}$ (stretched) and ${\textcolor{#2b7a2b}{y=\frac{1}{2}x^2}}$ (compressed).

If the factor in front of $f(x)$ is negative, for example ${y=-2x^2},$ the graph is stretched or compressed by the absolute value of the factor (here by $2$) and also reflected about the $x$-axis.

Common Mistakes

Mistake	Wrong	Right
Swapped coordinates	The point $(2, 5)$ is $2$ units up and $5$ units to the right	The $x$-coordinate comes first: $2$ units to the right and $5$ units up
A point on an axis	The point $(0, -4)$ lies in quadrant IV	It lies on the $y$-axis; points on the axes belong to no quadrant
Joining the branches	The graph of ${y=\frac{1}{x}}$ is one curve through the origin	${x\ne0}\text{:}$ the graph has two branches, in quadrants I and III
Forgetting the hole	The graph of ${y=\frac{x^2-4}{x-2}}$ is the whole line ${y=x+2}$	It is this line without the point $(2, 4)$
The direction of a horizontal shift	${y=(x-3)^2}$ is ${y=x^2}$ moved $3$ units to the left	It is moved $3$ units to the right
${f(x)+c}$ and ${f(x+c)}$	${y=x^2+3}$ is ${y=x^2}$ moved $3$ units to the left	It is moved $3$ units up; ${y=(x+3)^2}$ is moved to the left
The wrong line in the test	The parabola ${y=x^2}$ is not the graph of a function, because the line ${y=4}$ meets it twice	The test uses vertical lines; every vertical line meets the parabola ${y=x^2}$ exactly once

Practice Problems

1. In which quadrant or on which axis does each point lie? $A(3, -7),$ $B(-2, -5),$ $C(-4, 6),$ $D(0, 8),$ $E(5, 0).$

Answer

$A$ in quadrant IV, $B$ in quadrant III, $C$ in quadrant II, $D$ on the $y$-axis, $E$ on the $x$-axis.

2. The point $P(a, b)$ lies in quadrant II. In which quadrants do the points a) $(-a, b);$ b) $(a, -b);$ c) $(b, a)$ lie?

Answer

In quadrant II ${a\lt0}$ and ${b\gt0}.$ a) ${-a\gt0}$ and ${b\gt0}\text{:}$ quadrant I. b) ${a\lt0}$ and ${-b\lt0}\text{:}$ quadrant III. c) ${b\gt0}$ and ${a\lt0}\text{:}$ quadrant IV.

3. Which of the points $(1, 2),$ $(-2, 11)$ and $(3, 3)$ lie on the graph of ${y=x^2-2x+3}?$

Answer

${1-2+3=2}\text{:}$ yes. ${4+4+3=11}\text{:}$ yes. ${9-6+3=6\ne3}\text{:}$ no.

4. Find all intercepts of the graphs of a) ${2x-5y=10};$ b) ${y=x^2-9};$ c) ${y=x^2+1}.$

Answer

a) ${y=0}\text{:}$ ${2x=10},$ the $x$-intercept is $5.$ ${x=0}\text{:}$ ${-5y=10},$ the $y$-intercept is $-2.$

b) ${x^2=9}\text{:}$ the $x$-intercepts are $-3$ and $3;$ the $y$-intercept is $-9.$

c) ${x^2+1=0}$ has no real solution, so there is no $x$-intercept; the $y$-intercept is $1.$

5. Test the graphs for symmetry about the $x$-axis, the $y$-axis and the origin: a) ${y=x^4-3x^2};$ b) ${y=x^3-x};$ c) ${x^2+y^2=25}.$

Answer

a) Only about the $y$-axis: ${(-x)^4-3(-x)^2}={x^4-3x^2}.$

b) Only about the origin: ${-y=(-x)^3-(-x)}$ is the same as ${y=x^3-x}.$

c) About both axes and the origin (the graph is a circle with center $O$): $x$ and $y$ appear only squared.

6. Is the curve the graph of a function ${y=f(x)}?$ a) ${y=x^2-1};$ b) ${x=y^2};$ c) ${x^2+y^2=4}.$

Answer

a) Yes: for each $x$ there is exactly one $y.$

b) No: the vertical line ${x=4}$ meets the curve at $(4, 2)$ and $(4, -2).$

c) No: the vertical line ${x=0}$ meets the circle at $(0, 2)$ and $(0, -2).$

7. How is the graph obtained from the graph of ${y=x^2}$ or ${y=|x|}?$ a) ${y=(x+4)^2-1};$ b) ${y=|x-1|-3};$ c) ${y=-x^2+3}.$

Answer

a) ${y=x^2}$ moved $4$ units to the left and $1$ unit down; the vertex is $(-4, -1).$

b) ${y=|x|}$ moved $1$ unit to the right and $3$ units down; the corner is $(1, -3).$

c) ${y=x^2}$ reflected about the $x$-axis and moved $3$ units up; the vertex is $(0, 3).$

8. Complete the square in ${y=x^2+6x+7},$ find the vertex of the parabola and describe the translation of ${y=x^2}.$

Answer

${y}={(x^2+6x+9)-2}={(x+3)^2-2}.$ The vertex is $(-3, -2)\text{:}$ the parabola ${y=x^2}$ is moved $3$ units to the left and $2$ units down.

9. The graph of ${y=f(x)}$ passes through the point $(2, 5).$ Which point must lie on the graph of a) ${y=f(x)-4};$ b) ${y=f(x+1)};$ c) ${y=f(-x)};$ d) ${y=3f(x)}?$

Answer

a) $(2, 1);$ b) $(1, 5),$ because ${f(1+1)=f(2)=5};$ c) $(-2, 5);$ d) $(2, 15).$

10. Let ${h(x)=x^2}$ for ${x\lt1}$ and ${h(x)=3-x}$ for ${x\ge1}.$ Find $h(-2),$ $h(1)$ and $h(5).$ Which of the points $(1, 1)$ and $(1, 2)$ is on the graph, and how is each of them drawn in a sketch?

Answer

${h(-2)=4},$ ${h(1)=3-1=2},$ ${h(5)=-2}.$ The point $(1, 2)$ is on the graph (a filled dot). The left part of the graph ends at $(1, 1),$ which is not on the graph (an open circle).