The $x$-intercepts of the hyperbola with equation $\frac{x^{2}}{16}-\frac{y^{2}}{25}=1$ are $(4,0)$ and $(-4,0)$.

Name: The $x$-intercepts of the hyperbola with equation $\frac{x^{2}}{16}-\frac{y^{2}}{25}=1$ are $(4,0)$ and $(-4,0)$.
Brand: Quizplus
Rating: 4.8 (34 reviews)

Question

Quizplus · Accepted Answer

The $x$-intercepts occur when $y=0$. Substituting $y=0$ into the equation $\frac{x^{2}}{16}-\frac{y^{2}}{25}=1$, we get $\frac{x^{2}}{16}=1$, which simplifies to $x^{2}=16$. Solving for $x$, we find $x=4$ and $x=-4$, giving the points $(4,0)$ and $(-4,0)$.

The xxx -Intercepts of the Hyperbola with Equation x216−y225=1\frac{x^{2}}{16}-\frac{y^{2}}{25}=116x2​−25y2​=1 Are (4,0)(4,0)(4,0) And (−4,0)(-4,0)(−4,0)

The $x$ -Intercepts of the Hyperbola with Equation $\frac{x^{2}}{16}-\frac{y^{2}}{25}=1$ Are $(4,0)$ And $(-4,0)$