The $x$-intercepts of the ellipse with equation $\frac{x^{2}}{4}+\frac{y^{2}}{36}=1$ are $(2,0)$ and $(-2,0)$.

Name: The $x$-intercepts of the ellipse with equation $\frac{x^{2}}{4}+\frac{y^{2}}{36}=1$ are $(2,0)$ and $(-2,0)$.
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Quizplus · Accepted Answer

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

The xxx -Intercepts of the Ellipse with Equation x24+y236=1\frac{x^{2}}{4}+\frac{y^{2}}{36}=14x2​+36y2​=1 Are (2,0)(2,0)(2,0) And (−2,0)(-2,0)(−2,0)

The $x$ -Intercepts of the Ellipse with Equation $\frac{x^{2}}{4}+\frac{y^{2}}{36}=1$ Are $(2,0)$ And $(-2,0)$