Let's work through the given equation step by step to find its intercepts, domain, range, and asymptotes.

The Equation:

$2x^2 - 3y^2 = 6$

First, let's rewrite the equation in standard form.

Standard Form:

Divide both sides by 6 to simplify the equation: $\frac{2x^2}{6} - \frac{3y^2}{6} = 1$ This simplifies to: $\frac{x^2}{3} - \frac{y^2}{2} = 1$

This is the standard form of a hyperbola centered at the origin $(0, 0)$ with its transverse axis along the x-axis.

1. Intercepts:

x-intercepts:

To find the x-intercepts, set $y = 0$ and solve for $x$: $\frac{x^2}{3} - \frac{(0)^2}{2} = 1 \quad \Rightarrow \quad \frac{x^2}{3} = 1 \quad \Rightarrow \quad x^2 = 3 \quad \Rightarrow \quad x = \pm \sqrt{3}$ So, the x-intercepts are at: $x = \pm \sqrt{3}, \, y = 0$ or the points $(\sqrt{3}, 0)$ and $(-\sqrt{3}, 0)$.

y-intercepts:

To find the y-intercepts, set $x = 0$ and solve for $y$: $\frac{(0)^2}{3} - \frac{y^2}{2} = 1 \quad \Rightarrow \quad - \frac{y^2}{2} = 1 \quad \Rightarrow \quad y^2 = -2$ Since $y^2 = -2$ has no real solutions, there are no y-intercepts.

2. Domain:

The equation $\frac{x^2}{3} - \frac{y^2}{2} = 1$ represents a hyperbola. For real values of $y$, the term $\frac{x^2}{3}$ must be greater than or equal to 1. Hence, the domain is: $|x| \geq \sqrt{3}$ So, the domain in interval notation is: $(-\infty, -\sqrt{3}] \cup [\sqrt{3}, \infty)$

3. Range:

For any real value of $x$ in the domain, $y^2$ can take on any non-negative value, so $y$ can take on any real value. Therefore, the range of the hyperbola is: $(-\infty, \infty)$

4. Asymptotes:

The asymptotes for a hyperbola in the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are given by: $y = \pm \frac{b}{a}x$ In our case, $a^2 = 3$ and $b^2 = 2$, so the asymptotes are: $y = \pm \frac{\sqrt{2}}{\sqrt{3}} x = \pm \frac{\sqrt{6}}{3} x$ Thus, the asymptotes are the lines: $y = \pm \frac{\sqrt{6}}{3} x$

Final Answer:

• Intercepts:
  • $x$-intercepts: $(\sqrt{3}, 0)$ and $(-\sqrt{3}, 0)$
  • No $y$-intercepts
• Domain: $(-\infty, -\sqrt{3}] \cup [\sqrt{3}, \infty)$
• Range: $(-\infty, \infty)$
• Asymptotes: $y = \pm \frac{\sqrt{6}}{3} x$

Would you like more details or have any questions?

Related Questions:

1. How do you find the center and axes of a hyperbola?
2. Can you explain why there are no y-intercepts for this hyperbola?
3. How do the asymptotes relate to the behavior of the hyperbola?
4. How do you convert between different forms of conic sections?
5. What is the geometric significance of the coefficients in the equation of a hyperbola?

Tip: For any hyperbola, its asymptotes help predict the shape and direction of its branches as they extend towards infinity.