$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

x squared over A squared minus y squared over b squared equals 1

If the x² term is positive, the hyperbola is horizontal.

If the x squared term is positive, the hyperbola is horizontal.

x-intercepts: (-a, 0) and (a,0) and no y-intercepts.

x-intercepts: (negative A comma 0) and (A,0) and no y-intercepts.

$\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$

y squared over b squared minus x squared over A squared equals 1

If the y² term is positive, the hyperbola is vertical.

If the y squared term is positive, the hyperbola is vertical.

y-intercepts: (0, -b) and (0, b) and no x-intercepts.

y-intercepts: (0, negative b) and (0, b) and no x-intercepts.

Asymptotes: $y=\pm \frac{b}{a}x$

Asymptotes: y equals plus minus b over A times x

Graph: $\frac{x^2}{16}-\frac{y^2}{9}=1$

Graph: x squared over 16 minus y squared over 9 equals 1

The x² term is positive, so the hyperbola is horizontal; in other words, its branches open to the left and to the right.

The x squared term is positive, so the hyperbola is horizontal; in other words, its branches open to the left and to the right.

$\frac{x^2}{16}-\frac{0^2}{9}=1$ → $\frac{x^2}{16}=1$ → $x^2=16$ → x = ± 4

x squared over 16 minus 0 squared over 9 equals 1rightward arrow x squared over 16 equals 1rightward arrow x squared equals 16rightward arrow x equals plus minus 4

Therefore the x-intercepts are (-4,0) and (4,0).

Therefore the x-intercepts are ( negative 4,0) and (4,0).

The x-intercepts are given by the positive and negative square roots of the denominator of the x² term.

The x-intercepts are given by the positive and negative squarede roots of the denominator of the x squared term.

a² = 16 and b² = 9, so a = 4 and b = 3.

A squared equals 16 and b squared equals 9, so A = 4 and b = 3.

The asymptotes, which are given by $y=\pm \frac{b}{a}x$ , are $y=\pm \frac{3}{4}x$.

The asymptotes, which are given by y equals plus minus b over A times x, are y equals plus minus three fourths times x.

Graph: $\frac{y^2}{25}-\frac{x^2}{9}=1$

Graph: y squared over 25 minus x squared over 9 equals 1

The y² term is positive, so the hyperbola is vertical; in other words, its branches open upward and downward.

The y squared term is positive, so the hyperbola is vertical; in other words, its branches open upward and downward.

$\frac{y^2}{25}-\frac{0^2}{9}=1$ → $\frac{y^2}{25}=1$ → $y^2=25$ → y = ± 5

y squared over 25 minus 0 squared over 9 equals 1 rightward arrow y squared over 25 equals 1 rightward arrow y squared equals 25 rightward arrow y equals plus minus 5

The y-intercepts are given by the positive and negative square roots of the denominator of the y² term.

The y-intercepts are given by the positive and negative square roots of the denominator of the y squared term.

a² = 9 and b² = 25, so a = 3 and b = 5.

A squared equals 9 and b squared equals 25, so A = 3 and b = 5

The asymptotes, which are given by $y=\pm \frac{b}{a}x$ , are $y=\pm \frac{5}{3}x$.

The asymptotes, which are given by y equals plus minus b over A times x, are y equals plus minus five thirds times x.

In the following exploration you will select values for a and b and then see the corresponding graph of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

In the following exploration you will select values for A and B and then see the corresponding graph of x squared over A squared minus y squared over b squared equals 1

You will find that all the hyperbolas generated in this exploration are horizontal because the x² term is positive.

You will find that all the hyperbolas generated in this exploration are horizontal because the x squared term is positive.

You will see how a², the denominator of the x², determines the x-intercepts and that there are no y-intercepts.

You will see how A squared, the denominator of the x squared, determines the x-intercepts and that there are no y-intercepts.

Finally, you will note that the asymptotes are lines that go through the origin with slopes of $\frac{b}{a}$ and $-\frac{b}{a}$.

Finally, you will note that the asymptotes are lines that go through the origin with slopes of b over A and negative b over A.

Graph: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Graph: x squared over A squared minus y squared over b squared equals 1

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

x squared over A squared minus y squared over b squared equals 1

In the following exploration you will select values for a and b and then see the corresponding graph of $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$.

In the following exploration you will select values for A and B and then see the corresponding graph of y squared over b squared minus x squared over A squared equals 1

You will find that all the hyperbolas generated in this exploration are vertical because the y² term is positive.

You will find that all the hyperbolas generated in this exploration are vertical because the y squared term is positive.

You will see how b², the denominator of the y² term, determines the y-intercepts and that there are no x-intercepts.

You will see how b squared, the denominator of the y squared term, determines the y-intercepts and that there are no x-intercepts.

Finally, you will note that the asymptotes are lines that go through the origin with slopes of $\frac{b}{a}$ and $-\frac{b}{a}$.

Finally, you will note that the asymptotes are lines that go through the origin with slopes of b over A and negative b over A

Graph: $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$

Graph: y squared over b squared minus x squared over A squared equals 1

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

y squared over b squared minus x squared over A squared equals 1

The Graph of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ or $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$ , where a>0 and b>0:

The Graph of x squared over A squared minus y squared over b squared equals 1 or y squared over b squared minus x squared over A squared equals 1, where A is larger than 0 and b is larger than 0: