To find the intercepts of the equation $y = 6x^2 - 6$, we need to determine both the y-intercept and the x-intercepts.

1. Finding the y-intercept:

The y-intercept occurs when $x = 0$. So, substitute $x = 0$ into the equation: $y = 6(0)^2 - 6 = -6$ Thus, the y-intercept is at $(0, -6)$.

2. Finding the x-intercepts:

The x-intercepts occur when $y = 0$. Set the equation equal to 0 and solve for $x$: $0 = 6x^2 - 6$ Add 6 to both sides: $6 = 6x^2$ Now, divide both sides by 6: $x^2 = 1$ Take the square root of both sides: $x = \pm 1$ So, the x-intercepts are $(1, 0)$ and $(-1, 0)$.

Conclusion:

• y-intercept: $(0, -6)$
• x-intercepts: $(1, 0)$ and $(-1, 0)$

Would you like further clarification on any of these steps?

Here are 5 related questions:

1. How do you find intercepts for more complex polynomial equations?
2. What is the general process for solving quadratic equations?
3. Can a parabola have more than two x-intercepts?
4. What does the symmetry of the parabola suggest about its intercepts?
5. How can you determine the vertex of a quadratic equation?

Tip: For quadratic equations, the discriminant $b^2 - 4ac$ helps determine the nature of the x-intercepts (real or imaginary).