Answer:
(-5, 0) and (5, 0)
Step-by-step explanation:
This equation fits the form for a hyperbola with x-intercepts. The standard form for such an equation is
[tex]\frac{x^2}{a^2}-\frac{y^2}{b^2}=1[/tex]
To get the equation in the question into this standard form, divide each term by 400.
[tex]\frac{16x^2}{400}-\frac{25y^2}{400}=\frac{400}{400}\\\frac{x^2}{25}-\frac{y^2}{16}=1[/tex]
To find the x-intercepts, make y = 0.
[tex]\frac{x^2}{25}=1\\x^2=25\\x=\pm 5[/tex]
The vertices are located at the points (-5, 0) and (5, 0).
Note: There are no y-intercepts; making x = 0 produces no real solutions for y.
