Create the equation of a hyperbola centered at the origin, with a horizontal transverse axis, vertex at (–7, 0), and asymptotes of y equals plus or minus six sevenths x period Show your work. (4 points)

So,
First, remember that the general equation of a hyperbola is given by the following:
[tex]\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1[/tex]In our problem, we're given that this hyperbola is centered at the origin, so C(h,k) = C(0,0). Then, h=0, and k=0.
Our equation can be written as:
[tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]The vertex of this hyperbola is set in the point (-7,0). We could use the following equation to find the value of a:
[tex]V=(h\pm a,k)[/tex]Where V is the vertex.
If we replace our values:
[tex](0\pm a,0)=(-7,0)[/tex]So,
[tex]a=\pm7[/tex]Since "a" is squared in the equation, it doesn't matter if it is -7 or 7. The +/- sign indicates that there are two vertices at (-7,0) and (7,0) respectively.
Now, to find b, we could use the fact that the asymptotes of the hyperbola are:
[tex]y=\pm\frac{6}{7}[/tex]Remember that these asymptotes have the form:
[tex]y=\pm\frac{b}{a}[/tex]So, if we compare:
[tex]\begin{gathered} a=\pm7 \\ b=\pm6 \end{gathered}[/tex]And, our equation will be:
[tex]\frac{x^2}{49}-\frac{y^2}{36}=1[/tex]