Hyperbolas

Graph a Hyperbola with Center at

The last conic section we will look at is called a hyperbola. We will see that the equation of a hyperbola looks the same as the equation of an ellipse, except it is a difference rather than a sum. While the equations of an ellipse and a hyperbola are very similar, their graphs are very different.

We define a hyperbola as all points in a plane where the difference of their distances from two fixed points is constant. Each of the fixed points is called a focus of the hyperbola.

The line through the foci, is called the transverse axis. The two points where the transverse axis intersects the hyperbola are each a vertex of the hyperbola. The midpoint of the segment joining the foci is called the center of the hyperbola. The line perpendicular to the transverse axis that passes through the center is called the conjugate axis. Each piece of the graph is called a branch of the hyperbola.

Again our goal is to connect the geometry of a conic with algebra. Placing the hyperbola on a rectangular coordinate system gives us that opportunity. In the figure, we placed the hyperbola so the foci \(((\text{-}c,0),(c,0))\) are on the x-axis and the center is the origin.

The definition states the difference of the distance from the foci to a point \((x,y)\) is constant. So \(|{d}_{1}-{d}_{2}|\) is a constant that we will call \(2a\) so \(|{d}_{1}-{d}_{2}|=2a.\) We will use the distance formula to lead us to an algebraic formula for an ellipse.

\(\begin{array}{llll} & & & \ |{d}_{1}\ -\ {d}_{2}|\ =2a \\ \text{Use the distance formula to find}\ {d}_{1},{d}_{2} & & & \ |\sqrt{{(x-(-c))}^{2}+{(y-0)}^{2}}-\sqrt{{(x-c)}^{2}+{(y-0)}^{2}}\ |=2a \\ \text{Eliminate the radicals.} & & & \\ \begin{array}{l}\text{To simplify the equation of the ellipse, we} \\ \text{let}\ {c}^{2}-{a}^{2}={b}^{2}.\end{array} & & & \ \frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{c}^{2}-{a}^{2}}=1\ \\ \begin{array}{l}\text{So, the equation of a hyperbola centered at} \\ \text{the origin in standard form is:}\end{array} & & & \ \frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1\ \end{array}\)

To graph the hyperbola, it will be helpful to know about the intercepts. We will find the x-intercepts and y-intercepts using the formula.

How to Graph a Hyperbola with Center

Try it.

Graph \(\frac{{x}^{2}}{25}-\frac{{y}^{2}}{4}=1.\)

Solution

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

Graph a Hyperbola with Center at

Hyperbolas are not always centered at the origin. When a hyperbola is centered at \((h,k)\) the equations changes a bit as reflected in the table.

Standard Forms of the Equation a Hyperbola with Center \((h,k)\)
	\(\frac{{(x-h)}^{2}}{{a}^{2}}-\frac{{(y-k)}^{2}}{{b}^{2}}=1\)	\(\frac{{(y-k)}^{2}}{{a}^{2}}-\frac{{(x-h)}^{2}}{{b}^{2}}=1\)
Orientation	Transverse axis is horizontal. Opens left and right	Transverse axis is vertical. Opens up and down
Center	\((h,k)\)	\((h,k)\)
Vertices	a units to the left and right of the center	a units above and below the center
Rectangle	Use a units left/right of center b units above/ below the center	Use a units above/below the center b units left/right of center

How to Graph a Hyperbola with Center

Try it.

Graph \(\frac{{(x-1)}^{2}}{9}-\frac{{(y-2)}^{2}}{16}=1\)

Solution

We summarize the steps for easy reference.

Be careful as you identify the center. The standard equation has \(x-h\) and \(y-k\) with the center as \((h,k).\)

Example

Try it.

Graph \(\frac{{(y+2)}^{2}}{9}-\frac{{(x+1)}^{2}}{4}=1.\)

Solution

Since the \({y}^{2}\text{-}\)term is positive, the hyperbola opens up and down.
Find the center, \((h,k).\)	Center: \((-1,-2)\)
Find a, b.	\(a=3\) \(b=2\)
Sketch the rectangle that goes through the points 3 units above and below the center and 2 units to the left/right of the center. Sketch the asymptotes, the lines through the diagonals of the rectangle. Mark the vertices. Graph the branches.

Again, sometimes we have to put the equation in standard form as our first step.

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

Identify Conic Sections by their Equations

Now that we have completed our study of the conic sections, we will take a look at the different equations and recognize some ways to identify a conic by its equation. When we are given an equation to graph, it is helpful to identify the conic so we know what next steps to take.

To identify a conic from its equation, it is easier if we put the variable terms on one side of the equation and the constants on the other.

Conic	Characteristics of \({x}^{2}\text{-}\) and \({y}^{2}\text{-}\) terms	Example
Parabola	Either \({x}^{2}\) OR \({y}^{2}.\) Only one variable is squared.	\(x=3{y}^{2}-2y+1\)
Circle	\({x}^{2}\text{-}\) and \({y}^{2}\text{-}\) terms have the same coefficients	\({x}^{2}+{y}^{2}=49\)
Ellipse	\({x}^{2}\text{-}\) and \({y}^{2}\text{-}\) terms have the same sign, different coefficients	\(4{x}^{2}+25{y}^{2}=100\)
Hyperbola	\({x}^{2}\text{-}\) and \({y}^{2}\text{-}\) terms have different signs, different coefficients	\(25{y}^{2}-4{x}^{2}=100\)

Example

Try it.

Identify the graph of each equation as a circle, parabola, ellipse, or hyperbola.

ⓐ \(9{x}^{2}+4{y}^{2}+56y+160=0\) ⓑ \(9{x}^{2}-16{y}^{2}+18x+64y-199=0\) ⓒ \({x}^{2}+{y}^{2}-6x-8y=0\) ⓓ \(y=-2{x}^{2}-4x-5\)

Solution

ⓐ

	\(9{x}^{2}+4{y}^{2}+56y+160=0\)
The \({x}^{2}\)- and \({y}^{2}\)-terms have the same sign and different coefficients.	Ellipse

ⓑ

	\(9{x}^{2}-16{y}^{2}+18x+64y-199=0\)
The \({x}^{2}\)- and \({y}^{2}\)-terms have different signs and different coefficients.	Hyperbola

ⓒ

	\({x}^{2}+{y}^{2}-6x-8y=0\)
The \({x}^{2}\)- and \({y}^{2}\)-terms have the same coefficients.	Circle

ⓓ

	\(y=-2{x}^{2}-4x-5\)
Only one variable, \(x\), is squared.	Parabola

Key Concepts

• Hyperbola: A hyperbola is all points in a plane where the difference of their distances from two fixed points is constant.
  
  Each of the fixed points is called a focus of the hyperbola.
  The line through the foci, is called the transverse axis.
  The two points where the transverse axis intersects the hyperbola are each a vertex of the hyperbola.
  The midpoint of the segment joining the foci is called the center of the hyperbola.
  The line perpendicular to the transverse axis that passes through the center is called the conjugate axis.
  Each piece of the graph is called a branch of the hyperbola.
  
  

  Standard Forms of the Equation a Hyperbola with Center \((0,0)\)
  	\(\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1\)	\(\frac{{y}^{2}}{{a}^{2}}-\frac{{x}^{2}}{{b}^{2}}=1\)
  Orientation	Transverse axis on the x-axis. Opens left and right	Transverse axis on the y-axis. Opens up and down
  Vertices	\((\text{-}a,0),\) \((a,0)\)	\((0,\text{-}a),\) \((0,a)\)
  x-intercepts	\((\text{-}a,0),\) \((a,0)\)	none
  y-intercepts	none	\((0,\text{-}a)\), \((0,a)\)
  Rectangle	Use \((\pm a,0)\) \((0,\pm b)\)	Use \((0,\pm a)\) \((\pm b,0)\)
  asymptotes	\(y=\frac{b}{a}x,\) \(y=-\frac{b}{a}x\)	\(y=\frac{a}{b}x,\) \(y=-\frac{a}{b}x\)

• How to graph a hyperbola centered at \((0,0).\)
  1. Write the equation in standard form.
  2. Determine whether the transverse axis is horizontal or vertical.
  3. Find the vertices.
  4. Sketch the rectangle centered at the origin intersecting one axis at \(\pm a\) and the other at \(\pm b.\)
  5. Sketch the asymptotes, the lines through the diagonals of the rectangle.
  6. Draw the two branches of the hyperbola.
  
  

  Standard Forms of the Equation a Hyperbola with Center \((h,k)\)
  	\(\frac{{(x-h)}^{2}}{{a}^{2}}-\frac{{(y-k)}^{2}}{{b}^{2}}=1\)	\(\frac{{(y-k)}^{2}}{{a}^{2}}-\frac{{(x-h)}^{2}}{{b}^{2}}=1\)
  Orientation	Transverse axis is horizontal. Opens left and right	Transverse axis is vertical. Opens up and down
  Center	\((h,k)\)	\((h,k)\)
  Vertices	a units to the left and right of the center	a units above and below the center
  Rectangle	Use a units left/right of center b units above/below the center	Use a units above/below the center b units left/right of center

• How to graph a hyperbola centered at \((h,k).\)
  1. Write the equation in standard form.
  2. Determine whether the transverse axis is horizontal or vertical.
  3. Find the center and \(a,b.\)
  4. Sketch the rectangle centered at \((h,k)\) using \(a,b.\)
  5. Sketch the asymptotes, the lines through the diagonals of the rectangle. Mark the vertices.
  6. Draw the two branches of the hyperbola.
  
  

  Conic	Characteristics of \({x}^{2}\text{-}\) and \({y}^{2}\text{-}\) terms	Example
  Parabola	Either \({x}^{2}\) OR \({y}^{2}.\) Only one variable is squared.	\(x=3{y}^{2}-2y+1\)
  Circle	\({x}^{2}\text{-}\) and \({y}^{2}\text{-}\) terms must have the same coefficients and they must be the same sign as the constant after the = sign	\({x}^{2}+{y}^{2}=49\)
  Ellipse	\({x}^{2}\text{-}\) and \({y}^{2}\text{-}\) terms have the same sign, different coefficients	\(4{x}^{2}+25{y}^{2}=100\)
  Hyperbola	\({x}^{2}\text{-}\) and \({y}^{2}\text{-}\) terms have different signs	\(25{y}^{2}-4{x}^{2}=100\)

Hyperbolas

Graph a Hyperbola with Center at \((0,0)\)

In the following exercises, graph.

Try it.

\(\frac{{x}^{2}}{9}-\frac{{y}^{2}}{4}=1\)

Solution

Try it.

\(\frac{{x}^{2}}{25}-\frac{{y}^{2}}{9}=1\)

Try it.

\(\frac{{x}^{2}}{16}-\frac{{y}^{2}}{25}=1\)

Solution

Try it.

\(\frac{{x}^{2}}{9}-\frac{{y}^{2}}{36}=1\)

Try it.

\(\frac{{y}^{2}}{25}-\frac{{x}^{2}}{4}=1\)

Solution

Try it.

\(\frac{{y}^{2}}{36}-\frac{{x}^{2}}{16}=1\)

Try it.

\(16{y}^{2}-9{x}^{2}=144\)

Solution

Try it.

\(25{y}^{2}-9{x}^{2}=225\)

Try it.

\(4{y}^{2}-9{x}^{2}=36\)

Solution

Try it.

\(16{y}^{2}-25{x}^{2}=400\)

Try it.

\(4{x}^{2}-16{y}^{2}=64\)

Solution

Try it.

\(9{x}^{2}-4{y}^{2}=36\)

Graph a Hyperbola with Center at \((h,k)\)

In the following exercises, graph.

Try it.

\(\frac{{(x-1)}^{2}}{16}-\frac{{(y-3)}^{2}}{4}=1\)

Solution

Try it.

\(\frac{{(x-2)}^{2}}{4}-\frac{{(y-3)}^{2}}{16}=1\)

Try it.

\(\frac{{(y-4)}^{2}}{9}-\frac{{(x-2)}^{2}}{25}=1\)

Solution

Try it.

\(\frac{{(y-1)}^{2}}{25}-\frac{{(x-4)}^{2}}{16}=1\)

Try it.

\(\frac{{(y+4)}^{2}}{25}-\frac{{(x+1)}^{2}}{36}=1\)

Solution

Try it.

\(\frac{{(y+1)}^{2}}{16}-\frac{{(x+1)}^{2}}{4}=1\)

Try it.

\(\frac{{(y-4)}^{2}}{16}-\frac{{(x+1)}^{2}}{25}=1\)

Solution

Try it.

\(\frac{{(y+3)}^{2}}{16}-\frac{{(x-3)}^{2}}{36}=1\)

Try it.

\(\frac{{(x-3)}^{2}}{25}-\frac{{(y+2)}^{2}}{9}=1\)

Solution

Try it.

\(\frac{{(x+2)}^{2}}{4}-\frac{{(y-1)}^{2}}{9}=1\)

In the following exercises, ⓐ write the equation in standard form and ⓑ graph.

Try it.

\(9{x}^{2}-4{y}^{2}-18x+8y-31=0\)

Solution

ⓐ \(\frac{{(x-1)}^{2}}{4}-\frac{{(y-1)}^{2}}{9}=1\)
ⓑ

Try it.

\(16{x}^{2}-4{y}^{2}+64x-24y-36=0\)

Try it.

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

Solution

ⓐ \(\frac{{(y-2)}^{2}}{9}-\frac{{(x-1)}^{2}}{9}=1\)
ⓑ

Try it.

\(4{y}^{2}-16{x}^{2}-24y+96x-172=0\)

Try it.

\(9{y}^{2}-{x}^{2}+18y-4x-4=0\)

Solution

ⓐ \(\frac{{(y+1)}^{2}}{1}-\frac{{(x+2)}^{2}}{9}=1\)
ⓑ

Identify the Graph of each Equation as a Circle, Parabola, Ellipse, or Hyperbola

In the following exercises, identify the type of graph.

Try it.

ⓐ \(x=\text{-}{y}^{2}-2y+3\) ⓑ \(9{y}^{2}-{x}^{2}+18y-4x-4=0\) ⓒ \(9{x}^{2}+25{y}^{2}=225\) ⓓ \({x}^{2}+{y}^{2}-4x+10y-7=0\)

Try it.

ⓐ \(x=-2{y}^{2}-12y-16\) ⓑ \({x}^{2}+{y}^{2}=9\) ⓒ \(16{x}^{2}-4{y}^{2}+64x-24y-36=0\) ⓓ \(16{x}^{2}+36{y}^{2}=576\)

Solution

ⓐ parabola ⓑ circle ⓒ hyperbola ⓓ ellipse

Mixed Practice

In the following exercises, graph each equation.

Try it.

\(\frac{{(y-3)}^{2}}{9}-\frac{{(x+2)}^{2}}{16}=1\)

Try it.

\({x}^{2}+{y}^{2}-4x+10y-7=0\)

Solution

Try it.

\(y={(x-1)}^{2}+2\)

Try it.

\(\frac{{x}^{2}}{9}+\frac{{y}^{2}}{25}=1\)

Solution

Try it.

\({(x+2)}^{2}+{(y-5)}^{2}=4\)

Try it.

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

Solution

Try it.

\(x=\text{-}{y}^{2}-2y+3\)

Try it.

\(16{x}^{2}+9{y}^{2}=144\)

Solution

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

Dabar tu Joks skaičiuotuvas nenusistato, bet jo dalys yra skaičiuojamos. Pabandykite vieną žemiau arba įveskite savo.

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Jūsų matematikos problema

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Praktika (40)

Išbandykite kiekvieną ant popieriaus pirmą. Atskleisti atsakymą patikrinti; patikrintas tie gali būti atidarytas sprendėjas už kiekvieną žingsnį.

1. Solve: \({x}^{2}=12.\)

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   \(x=\pm 2\sqrt{3}\)

2. Expand: \({(x-4)}^{2}.\)

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   \({x}^{2}-8x+16\)

3. Graph \(y=-\frac{2}{3}x.\)

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4. Graph \(\frac{{x}^{2}}{25}-\frac{{y}^{2}}{4}=1.\)

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5. Graph \(\frac{{x}^{2}}{16}-\frac{{y}^{2}}{4}=1.\)

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6. Graph \(\frac{{x}^{2}}{9}-\frac{{y}^{2}}{16}=1.\)

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7. Graph \(4{y}^{2}-16{x}^{2}=64.\)

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   	\(4{y}^{2}-16{x}^{2}=64\)
   To write the equation in standard form, divide each term by 64 to make the equation equal to 1.	\(\frac{4{y}^{2}}{64}-\frac{16{x}^{2}}{64}=\frac{64}{64}\)
   Simplify.	\(\ \frac{{y}^{2}}{16}-\frac{{x}^{2}}{4}=1\)
   Since the y2-term is positive, the transverse axis is vertical. Since \({a}^{2}=16\) then \(a=\pm 4.\)
   The vertices are on the y-axis, \((0,\text{-}a),\) \((0,a).\) Since \({b}^{2}=4\) then \(b=\pm 2.\)	\((0,-4),\) \((0,4)\)
   Sketch the rectangle intersecting the x-axis at \((-2,0),\) \((2,0)\) and the y-axis at the vertices. Sketch the asymptotes through the diagonals of the rectangle. Draw the two branches of the hyperbola.

8. Graph \(4{y}^{2}-25{x}^{2}=100.\)

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9. Graph \(25{y}^{2}-9{x}^{2}=225.\)

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10. Graph \(\frac{{(x-1)}^{2}}{9}-\frac{{(y-2)}^{2}}{16}=1\)

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11. Graph \(\frac{{(x-3)}^{2}}{25}-\frac{{(y-1)}^{2}}{9}=1.\)

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12. Graph \(\frac{{(x-2)}^{2}}{4}-\frac{{(y-2)}^{2}}{9}=1.\)

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13. Graph \(\frac{{(y+2)}^{2}}{9}-\frac{{(x+1)}^{2}}{4}=1.\)

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    Atskleisti atsakymą

    Since the \({y}^{2}\text{-}\)term is positive, the hyperbola opens up and down.
    Find the center, \((h,k).\)	Center: \((-1,-2)\)
    Find a, b.	\(a=3\) \(b=2\)
    Sketch the rectangle that goes through the points 3 units above and below the center and 2 units to the left/right of the center. Sketch the asymptotes, the lines through the diagonals of the rectangle. Mark the vertices. Graph the branches.

14. Graph \(\frac{{(y+3)}^{2}}{16}-\frac{{(x+2)}^{2}}{9}=1.\)

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15. Graph \(\frac{{(y+2)}^{2}}{9}-\frac{{(x+2)}^{2}}{9}=1.\)

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16. Write the equation in standard form and graph \(4{x}^{2}-9{y}^{2}-24x-36y-36=0.\)

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    Atskleisti atsakymą

    To get to standard form, complete the squares.
    Divide each term by 36 to get the constant to be 1.
    Since the \({x}^{2}\text{-}\)term is positive, the hyperbola opens left and right.
    Find the center, \((h,k).\)	Center: \((3,-2)\)
    Find a, b.	\(\begin{array}{l}a=3 \\ b=4\end{array}\)
    Sketch the rectangle that goes through the points 3 units to the left/right of the center and 2 units above and below the center. Sketch the asymptotes, the lines through the diagonals of the rectangle. Mark the vertices. Graph the branches.

17. ⓐ Write the equation in standard form and ⓑ graph \(9{x}^{2}-16{y}^{2}+18x+64y-199=0.\)

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    Atskleisti atsakymą

    ⓐ \(\frac{{(x+1)}^{2}}{16}-\frac{{(y-2)}^{2}}{9}=1\)
    ⓑ

18. ⓐ Write the equation in standard form and ⓑ graph \(16{x}^{2}-25{y}^{2}+96x-50y-281=0.\)

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    Atskleisti atsakymą

    ⓐ \(\frac{{(x+3)}^{2}}{25}-\frac{{(y+1)}^{2}}{16}=1\)
    ⓑ

19. Identify the graph of each equation as a circle, parabola, ellipse, or hyperbola.

    ⓐ \(9{x}^{2}+4{y}^{2}+56y+160=0\) ⓑ \(9{x}^{2}-16{y}^{2}+18x+64y-199=0\) ⓒ \({x}^{2}+{y}^{2}-6x-8y=0\) ⓓ \(y=-2{x}^{2}-4x-5\)

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    ⓐ
    

    	\(9{x}^{2}+4{y}^{2}+56y+160=0\)
    The \({x}^{2}\)- and \({y}^{2}\)-terms have the same sign and different coefficients.	Ellipse

    
    
    ⓑ
    

    	\(9{x}^{2}-16{y}^{2}+18x+64y-199=0\)
    The \({x}^{2}\)- and \({y}^{2}\)-terms have different signs and different coefficients.	Hyperbola

    
    
    ⓒ
    

    	\({x}^{2}+{y}^{2}-6x-8y=0\)
    The \({x}^{2}\)- and \({y}^{2}\)-terms have the same coefficients.	Circle

    
    
    ⓓ
    

    	\(y=-2{x}^{2}-4x-5\)
    Only one variable, \(x\), is squared.	Parabola

20. Identify the graph of each equation as a circle, parabola, ellipse, or hyperbola.

    ⓐ \({x}^{2}+{y}^{2}-8x-6y=0\) ⓑ \(4{x}^{2}+25{y}^{2}=100\) ⓒ \(y=6{x}^{2}+2x-1\) ⓓ \(16{y}^{2}-9{x}^{2}=144\)

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    Atskleisti atsakymą

    ⓐ circle ⓑ ellipse ⓒ parabola ⓓ hyperbola

21. Identify the graph of each equation as a circle, parabola, ellipse, or hyperbola.

    ⓐ \(16{x}^{2}+9{y}^{2}=144\) ⓑ \(y=2{x}^{2}+4x+6\) ⓒ \({x}^{2}+{y}^{2}+2x+6y+9=0\) ⓓ \(4{x}^{2}-16{y}^{2}=64\)

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    Atskleisti atsakymą

    ⓐ ellipse ⓑ parabola ⓒ circle ⓓ hyperbola

22. \(\frac{{x}^{2}}{9}-\frac{{y}^{2}}{4}=1\)

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23. \(\frac{{x}^{2}}{25}-\frac{{y}^{2}}{9}=1\)

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24. \(\frac{{x}^{2}}{16}-\frac{{y}^{2}}{25}=1\)

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25. \(\frac{{x}^{2}}{9}-\frac{{y}^{2}}{36}=1\)

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26. \(\frac{{y}^{2}}{25}-\frac{{x}^{2}}{4}=1\)

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27. \(\frac{{y}^{2}}{36}-\frac{{x}^{2}}{16}=1\)

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28. \(16{y}^{2}-9{x}^{2}=144\)

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29. \(25{y}^{2}-9{x}^{2}=225\)

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30. \(4{y}^{2}-9{x}^{2}=36\)

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31. \(16{y}^{2}-25{x}^{2}=400\)

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32. \(4{x}^{2}-16{y}^{2}=64\)

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33. \(9{x}^{2}-4{y}^{2}=36\)

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34. \(\frac{{(x-1)}^{2}}{16}-\frac{{(y-3)}^{2}}{4}=1\)

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35. \(\frac{{(x-2)}^{2}}{4}-\frac{{(y-3)}^{2}}{16}=1\)

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36. \(\frac{{(y-4)}^{2}}{9}-\frac{{(x-2)}^{2}}{25}=1\)

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37. \(\frac{{(y-1)}^{2}}{25}-\frac{{(x-4)}^{2}}{16}=1\)

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38. \(\frac{{(y+4)}^{2}}{25}-\frac{{(x+1)}^{2}}{36}=1\)

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39. \(\frac{{(y+1)}^{2}}{16}-\frac{{(x+1)}^{2}}{4}=1\)

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40. \(\frac{{(y-4)}^{2}}{16}-\frac{{(x+1)}^{2}}{25}=1\)

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Išlaikyti savo darbą

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Čia naudojami simboliai

Bakstelėkite visus simbolius, kad būtų galima apibrėžti visą apibrėžimą, paveikslėlį ir ką reiškia kiekviena raidė.

\sqrt{x}square root, n-th root
The non-negative number whose square (n-th power) is x.\pmplus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.|x|absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.\binom{n}{k}binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).\pipi
Ratio of a circle's circumference to its diameter, 3.14159…eEuler's number
2.71828…, the base whose exponential is its own derivative.iimaginary unit
i² = −1.\thetatheta
The usual name for an angle.\log_b xlogarithm, natural log
The exponent b must be raised to for x; ln uses base e.\mathbf{v}vector
A quantity with magnitude and direction; a column of numbers.

Kaip vartoti: Hyperbolas

1. Graph a hyperbola with center at
2. Graph a hyperbola with center at
3. Identify conic sections by their equations
4. Write the equation in standard form.
5. Determine whether the transverse axis is horizontal or vertical.
6. Find the vertices.
7. Sketch the rectangle centered at the origin intersecting one axis at
8. Sketch the asymptotes—the lines through the diagonals of the rectangle.

Klausimai, kuriuos klausia žmonės

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

Šio puslapio dalys pritaikytos nuo OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Čia yra įtikinamų ir iš naujo paaiškintų klaidų.

Daugiau informacijos Precalculus

Complex numbersPolynomial functionsRational functionsSequences and seriesThe binomial theoremConic sectionsVectorsExponential and logarithmic functionsPolynomial division and the remainder theoremParametric equations and polar coordinates