OBJECTIVES

When you finish this section, you should be able to:

1. Use the distance formula (p. A-16)
2. Graph equations, find intercepts, and test for symmetry (p. A-16)
3. Work with equations of a line (p. A-18)
4. Work with the equation of a circle (p. A-21)
5. Graph parabolas, ellipses, and hyperbolas (p. A-22)

NOTE

If \((x, y)\) are the coordinates of a point \(P\), then \(x\) is called the \(x\)-coordinate , of \(P\) and \(y\) is called the \(y\)-coordinate , of \(P\).

THEOREM Distance Formula

The distance between two points \(P_{1}=( x_{1}, y_{1}) \) and \(P_{2}=( x_{2}, y_{2}) \), denoted by \(d( P_{1}, P_{1}) \), is \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ d ( P_{1}, P_{2}) =\displaystyle\sqrt{ ( x_{2}-x_{1}) ^{2}+ ( y_{2}-y_{1}) ^{2}} }} \]

Using the Distance Formula

Find the distance \(d\) between the points \(( -3,5)\) and \(( 3,2) .\)

Solution We use the distance formula with \(P_{1}= ( x_{1}, y_{1}) =( -3,5) \) and \(P_{2}= ( x_{2}, y_{2}) =( 3,2) .\) \[ d=\displaystyle\sqrt{ [ 3-( -3) ] ^{2}+( 2-5) ^{2}}=\displaystyle\sqrt{ 6^{2}+( -3) ^{2}}=\displaystyle\sqrt{36+9}=\displaystyle\sqrt{45}=3\displaystyle\sqrt{5}\approx 6.708 \]

Graphing an Equation by Plotting Points

Graph the equation \(y=x^{2}\).

Solution Table 4 lists several points on the graph. In Figure 21(a) the points are plotted and then they are connected with a smooth curve to obtain the graph (a parabola). Figure 21(b) shows the graph using graphing technology.

Figure 21 \(y=x^2\)

The graphs in Figure 21 do not show all the points whose coordinates \(( x,y)\) satisfy the equation \(y=x^{2}\). For example, the point \(( 6, 36)\) satisfies the equation \(y=x^{2}\), but it is not shown. It is important when graphing to present enough of the graph so that any viewer will “see” the rest of the graph as an obvious continuation of what is actually there.

Table 4

Figure 22

The points, if any, at which a graph crosses or touches a coordinate axis are called the intercepts of the graph. See Figure 22. The \(x\)-coordinate of a point where a graph crosses or touches the \(x\)-axis is an \( x\) -intercept. At an \(x\)-intercept, the \( y\)-coordinate equals \( 0.\) The \(y\)-coordinate of a point where a graph crosses or touches the \(y\) -axis is a \( y\)-intercept. At a \(y\)-intercept, the \(x\)-coordinate equals \(0\). When graphing an equation, all its intercepts should be displayed or be easily inferred from the part of the graph that is displayed.

Finding the Intercepts of a Graph

Find the \(x\)-intercept(s) and the \(y\)-intercept(s) of the graph of \( y=x^{2}-4 \).

Solution  To find the \(x\)-intercept(s), we let \(y=0\) and solve the equation \[ \begin{array}{rl@{{16pt}}l} x^{2}-4=0 & & \\[3pt] ( x+2) ( x-2) =0 & & {\color{#0066A7}{\hbox{Factor.}}} \\[3pt] x=-2{\rm \quad or \quad } x=2 & & {\color{#0066A7}{\hbox{Set each factor equal to 0 and solve.}}} \end{array} \]

The equation has two solutions, \(-2\) and \(2\). The \(x\)-intercepts are \(-2\) and \(2\).

To find the \(y\)-intercept(s), we let \(x=0\) in the equation. \[ y=0^{2}-4=-4 \]

The \(y\)-intercept is \(-4\).

spanDEFINITIONspan Symmetry

• A graph is symmetric with respect to the \({x }\) -axis if, for every point \(( x, y)\) on the graph, the point \(( x,-y)\) is also on the graph. See Figure 23.
• A graph is symmetric with respect to the \({y }\) -axis if, for every point \(( x, y)\) on the graph, the point \(( -x, y)\) is also on the graph. See Figure 24.
• A graph is symmetric with respect to the origin if, for every point \(( x, y)\) on the graph, the point \(( -x,-y)\) is also on the graph. See Figure 25.

Testing for Symmetry

Test the graph of \(y=\dfrac{4x^{2}}{x^{2}+1}\) for symmetry.

Solution  \(x\)-axis: To test for symmetry with respect to the \( x \)-axis, we replace \(y\) with \(-y\). Since \(-y=\dfrac{4x^{2}}{x^{2}+1}\) is not equivalent to \(y=\dfrac{4x^{2}}{x^{2}+1}\), the graph of the equation is not symmetric with respect to the \(x\)-axis.

\(y\)-axis: To test for symmetry with respect to the \(y\)-axis, we replace \(x\) with \(-x\). Since \(y=\dfrac{4( -x) ^{2}}{( -x) ^{2}+1}= \dfrac{4x^{2}}{x^{2}+1}\) is equivalent to \(y=\dfrac{4x^{2}}{x^{2}+1}\), the graph of the equation is symmetric with respect to the \(y\)-axis.

Origin: To test for symmetry with respect to the origin, we replace \(x\) with \( -x\) and \(y\) with \(-y\). \begin{eqnarray*} -y &=&\dfrac{4( -x) ^{2}}{( -x) ^{2}+1}\quad{\color{#0066A7}{\hbox{Replace x by -x and y by -y.}}} \\[1pt] -y &=&\dfrac{4x^{2}}{x^{2}+1}\quad {\color{#0066A7}{\hbox{Simplify.}}} \\[1pt] y &=&-\dfrac{4x^{2}}{x^{2}+1}\quad{\color{#0066A7}{\hbox{Multiply both sides by -1.}}} \end{eqnarray*}

spanDEFINITIONspan Slope

Let \(P= ( x_{1}, y_{1})\) and \(Q= ( x_{2}, y_{2})\) be two distinct points. If \(x_{1}\,{\neq}\,x_{2}\), the slope \(m\) of the nonvertical line \(L\) containing \(P\) and \(Q\) is defined by the formula \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}} \qquad x_{1}\neq x_{2} }} \]

If \(x_{1}=x_{2}\), then \(L\) is a vertical line and the slope \(m\) of \(L\) is undefined.

THEOREM Equation of a Vertical Line

A vertical line is given by an equation of the form \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ x=a}} \] where \(a\) is the \(x\)-intercept.

THEOREM Point-Slope Form of an Equation of a Line

An equation of a nonvertical line with slope \(m\) that contains the point\( ( x_{1},y_{1})\) is \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ y-y_{1}=m\,( x-x_{1}) }}\]

Using the Point-Slope Form of a Line

An equation of the line with slope \(4\) and containing the point \(( 1,2)\) can be found by using the point-slope form with \(m=4\), \(x_{1}=1\), and \(y_{1}=2\). \[ \begin{array}{lrcll@{{16pt}}ll} & y-y_{1}&=&m( x-x_{1}) & & {\color{#0066A7}{\hbox{Point-slope form of a line}}} & \\[3pt] & y-2&=&4( x-1) & & {\color{#0066A7}{{m=4}\hbox{, }{x}_{1}{=1} \hbox{, }{y}_{1}{=2}}} & \\[3pt] \,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ & ~\ \ \ \ \ y&=&4x-2 & & {\color{#0066A7}{\hbox{ Solve for }{y.}}} & \hbox{ } \end{array} \]

Figure 28 \(y=4x-2\)

See Figure 28 for the graph of \(y\).

THEOREM Slope-Intercept Form of an Equation of a Line

An equation of a line with slope \(m\) and \(y\)-intercept \(b\) is \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ y=mx+b }} \]

For a horizontal line, the slope \(m\) is \(0\).

THEOREM Equation of a Horizontal Line

A horizontal line is given by an equation of the form \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ y=b }} \] where \(b\) is the \(y\)-intercept.

Finding the Slope and y-intercept from an Equation of a Line

Find the slope \(m\) and the \(y\)-intercept of the equation \(2x+4y=8\). Graph the equation.

Solution  To obtain the slope and \(y\)-intercept, we write the equation in slope-intercept form by solving for \(y\). \begin{eqnarray*} 2x+4y &=&8 \\[3pt] 4y &=&-2x+8 \\[3pt] y &=&-\dfrac{1}{2}x+2 \quad{\color{#0066A7}{y=mx+b}} \end{eqnarray*}

The coefficient of \(x\) is the slope. The slope is \(-\dfrac{1}{2}.\) The constant \(2\) is the \(y\)-intercept, so the point \(( 0,2)\) is on the graph. Now use the slope \(-\dfrac{1}{2}.\) Starting at the point \(( 0,2)\), we move \(2\) units to the right and then \(1\) unit down to the point \((2, 1)\). We plot this point and draw a line through the two points. See Figure 29.

Figure 29 \(2x+4y=8\)

When two lines in the plane do not intersect, they are parallel.

THEOREM Criterion for Parallel Lines

Two nonvertical lines are parallel if and only if their slopes are equal and they have different \(y\)-intercepts.

Finding an Equation of a Line that Is Parallel to a Given Line

Find an equation of the line that contains the point \((1,2) \) and is parallel to the line \(y=5x\).

Solution  Since the two lines are parallel, the slope of the line we seek equals the slope of the line \(y=5x,\) which is \(5.\) Since the line we seek also contains the point \(( 1,2) \), we use the point-slope form to obtain the equation. \begin{eqnarray*} y-y_{1} &=&m( x-x_{1}) \\[0pt] y-2 &=&5( x-1) \\[0pt] y-2 &=&5x-5 \\[0pt] y &=&5x-3 \end{eqnarray*}

The line \(y=5x-3\) contains the point \(( 1,2)\) and is parallel to the line \(y=5x.\) See Figure 30.

THEOREM Criterion for Perpendicular Lines

Two nonvertical lines are perpendicular if and only if the product of their slopes is \(-1.\)

Finding an Equation of a Line that Is Perpendicular to a Given Line

Find an equation of the line that contains the point \(( -1,3)\) and is perpendicular to the line \(4x+y=-1.\)

Solution  We begin by writing the equation of the given line in slope-intercept form to find its slope. \begin{eqnarray*} 4x+y &=&-1 \\[0pt] y &=&-4x-1 \end{eqnarray*}

Figure 32

This line has a slope of \(-4.\) Any line perpendicular to this line will have slope \(\dfrac{1}{4}.\) Because the point \( ( -1,3)\) is on this line, we use the point-slope form of a line. \begin{eqnarray*} y-y_{1} &=&m( x-x_{1}) \\[3pt] y-3 &=&\dfrac{1}{4} [ x- ( -1) ] \\[4pt] y-3 &=&\dfrac{1}{4}x+\dfrac{1}{4} \\[4pt] y &=&\dfrac{1}{4}x+\dfrac{13}{4} \end{eqnarray*}

Figure 32 shows the graphs of \(4x+y=-1\) and \(y=\dfrac{1}{4}x+\dfrac{13}{4}\).

spanDEFINITIONspan Circle

A circle is the set of points in the \(xy\)-plane that is a fixed distance \(r\) from a fixed point \(( h,\,k) \). The fixed distance \(r\) is called the radius, and the fixed point \(( h,\,k)\) is called the center of the circle.

THEOREM

The standard form of a circle with radius \(r\) and center at the origin \(( 0,0)\) is \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ x^{2}+y^{2}=r^{2}} } \]

spanDEFINITIONspan Unit Circle

The circle with radius \(r=1\) and center at the origin is called the unit circle and has the equation \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ x^{2}+y^{2}=1 }}\]

Graphing a Circle

Graph the equation \(( x+3) ^{2}+( y-2) ^{2}=16\).

Solution  This is the standard form of an equation of a circle. To graph the circle, we first identify the center and the radius of the circle. \begin{eqnarray*} ( x+3) ^{2}+( y-2) ^{2}&=&16\\[0pt] ( x-( \underset{\underset{\color{#0066A7}{h}}{\color{#0066A7}{\uparrow}}}{-3}))^{2} +( y-\underset{\underset{\color{#0066A7}{k}}{\color{#0066A7}{\uparrow}}}{2})^{2}&=&\underset{\underset{\color{#0066A7}{r^2}}{\color{#0066A7}{\uparrow}}}{4^{2}}\\[0pt] \end{eqnarray*}

Figure 35 \(( x+3) ^{2}+( y-2) ^{2}=16\)

The circle has its center at the point\( ( -3, 2) ;\) its radius is \(4\) units. To graph the circle, we plot the center \(( -3, 2) .\) Then we locate the four points on the circle that are \(4\) units to the left, to the right, above, and below the center. These four points are used as guides to obtain the graph. See Figure 35.

spanDEFINITIONspan General Form of the Equation of a Circle

The equation \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ x^{2}+y^{2}+ax+bx+c=0 } }\] is called the general form of the equation of a circle.

NEED TO REVIEW?

Completing the square is discussed in Section A.1, p. A-2.

Graphing a Circle Whose Equation Is in General Form

Graph the equation \[ x^{2}+y^{2}+4x-6y+12=0 \]

Solution  We complete the square in both \(x\) and \(y\) to put the equation in standard form. First, we group the terms involving \(x\), group the terms involving \(y,\) and put the constant on the right side of the equation. The result is \[ ( x^{2}+4x) + ( y^{2}-6y) =-12 \]

Figure 36 \(x^{2}+y^{2}+4x-6y+12=0\)

Next, we complete the square of each expression in parentheses. Remember that any number added to the left side of an equation must be added to the right side. \begin{eqnarray*} {( x^{2}+4x\underset{\underset{\color{#0066A7}{\hbox{Add \(\left( \dfrac{4}{2}\right) ^{\!\!\!2}=\, 4\)}}}{\color{#0066A7}{\displaystyle\uparrow}}}{+}4) + ( y^{2}-6y\underset{\underset{\color{#0066A7}{\hbox{Add \(\left( \dfrac{-6}{2}\right) ^{\!\!\!2}=\, 9\)}}} {\color{#0066A7}{\displaystyle \uparrow}}}{+}9) }&=&-12+4+9=1\\ ( x+2)^{2}+( y-3) ^{2}&=&1 {16pt}\quad {\color{#0066A7}{\hbox{Factor.}}} \end{eqnarray*}

This is the standard form of the equation of a circle with radius \(1\) and center at the point \(( -2,3)\). To graph the circle, we plot the center \(( -2,3)\) and points \(1\) unit to the right, to the left, above and below the point \(( -2,3)\), as shown in Figure 36.

Graphing a Parabola with Vertex at the Origin

Graph the equation \(x^{2}=16y\).

Solution  The graph of \(x^{2}=16y\) is a parabola of the form \[ x^{2}=4ay \]

where \(a=4\). Look at Figure 37(c). The graph will open up, with focus at\( ( 0,4) \). The vertex is at \(( 0,0) \). To graph the parabola, we let \(y=a;\) that is, let \(y=4\). (Any other positive number will also work.) \begin{eqnarray*} x^{2} &=&16y \underset{\underset{\color{#0066A7}{\hbox{\(y=a=4\)}}}{\color{#0066A7}{\displaystyle\uparrow }}}{=} 16( 4) =64 \\[-5.6pt] x &=&-8\qquad\hbox{ or }\qquad x=8 \end{eqnarray*}

Figure 38 \(x^{2}=16y\)

The points\( ( -8,4)\) and\( ( 8,4)\) are on the parabola and establish its opening. See Figure 38.

NOTE

If \({a=b}\), then \(\dfrac{{x}^{2}}{ {a}^{2}}{+\dfrac{y^{2}}{a^{2}}=1}\) is the equation of a circle of radius \(a\).

Graphing an Ellipse with Center at the Origin

Graph the equation: \(9x^{2}+4y^{2}=36\)

Solution  To put the equation in standard form, we divide each side by \(36\). \[ \dfrac{x^{2}}{4}+\dfrac{y^{2}}{9}=1 \]

The graph of this equation is an ellipse. Since the larger number is under \( y^{2}\), the major axis is along the \(y\)-axis. The center is at the origin (0, 0). It is easiest to graph an ellipse by finding its intercepts: \[ \begin{array}{rcl@{\qquad\qquad\qquad\qquad}rcl} {x\hbox{-intercepts: Let }y=0} & {y\hbox{-intercepts: Let }x=0} \\[5pt] \dfrac{x^{2}}{4}+\dfrac{0^{2}}{9} &=& 1 & 0^{2}+\dfrac{y^{2}}{9}&=&1 \\[9pt] x^{2}&=&4 & y^{2}&=&9 \\[5pt] {x=-2 \hbox{ and } x=2} & {y=-3 \hbox{ and } y=3} \end{array} \]

Figure 40 \(9x^2 + 4y^2 = 36\)

The points \(( -2,0) \), \(( 2,0) \), \(( 0,-3) \) , \(( 0,3)\) are the intercepts of the ellipse. See Figure 40.

Graphing a Hyperbola With Center at the Origin

Graph the equation \(\dfrac{y^{2}}{4}-\dfrac{x^{2}}{5}=1\).

Solution  The graph of \(\dfrac{y^{2}}{4}-\dfrac{x^{2}}{5}=1\) is a hyperbola. The hyperbola consists of two branches, one opening up, the other opening down, like the graph in Figure 41(b). The hyperbola has no \(x\) -intercepts. To find the \(y\)-intercepts, we let \(x=0\) and solve for \(y.\) \[ \begin{array}{l} \dfrac{y^{2}}{4}=1 \\[11pt] y^{2}=4 \\[5pt] y=-2\qquad\hbox{ or} \qquad y=2 \end{array} \]

The \(y\)-intercepts are \(-2\) and \(2\), so the vertices are \((0,-2)\) and (\(0,2\)). The transverse axis is the vertical line \(x=0.\) To graph the hyperbola, let \(y=\pm\, 3\) (or any numbers \(\geq\)\(2\) or \(\leq\)\(-2\)). Then \[ \begin{array}{rclll} \dfrac{y^{2}}{4}-\dfrac{x^{2}}{5}&=&1 & & \\ \dfrac{9}{4}-\dfrac{x^{2}}{5}&=&1 & & {\color{#0066A7}{y=\pm\, 3}} \\ ~\ \ \ \ \ \ \dfrac{x^{2}}{5}&=&\dfrac{5}{4} & & \\ ~\ \ \ \ \ \ \ x^{2}&=&\dfrac{25}{4} & & \\ x=-\dfrac{5}{2} &\hbox{or}& x=\dfrac{5}{2} \\ \end{array} \]

Figure 42 \(\dfrac{y^{2}}{4}-\dfrac{x^{2}}{5}=1\)

The points \(\left( -\dfrac{5}{2},3\right) \), \(\left( -\dfrac{5}{2},-3\right) \), \(\left( \dfrac{5}{2},3\right) \), and \(\left( \dfrac{5}{2},-3\right)\) are on the hyperbola. See Figure 42 for the graph.