To find the intercepts of the given equation $y = x^2 - 6x - 27$, we'll determine the points where the graph of the equation crosses the x-axis and the y-axis.

1. Finding the x-intercepts

The x-intercepts occur where $y = 0$. To find these, set the equation equal to 0 and solve for $x$:

$x^2 - 6x - 27 = 0$

This is a quadratic equation. We can solve it using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For the equation $x^2 - 6x - 27 = 0$, the coefficients are:

• $a = 1$
• $b = -6$
• $c = -27$

Plugging these values into the quadratic formula:

$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-27)}}{2 \cdot 1}$

$x = \frac{6 \pm \sqrt{36 + 108}}{2}$

$x = \frac{6 \pm \sqrt{144}}{2}$

$x = \frac{6 \pm 12}{2}$

This gives two solutions:

$x_1 = \frac{6 + 12}{2} = \frac{18}{2} = 9$

$x_2 = \frac{6 - 12}{2} = \frac{-6}{2} = -3$

So, the x-intercepts are:

$(9, 0) \quad \text{and} \quad (-3, 0)$

2. Finding the y-intercept

The y-intercept occurs where $x = 0$. To find this, substitute $x = 0$ into the equation:

$y = 0^2 - 6 \cdot 0 - 27$

$y = -27$

So, the y-intercept is:

$(0, -27)$

3. Intercepts Summary

The intercepts are:

• x-intercepts: $(9, 0)$ and $(-3, 0)$
• y-intercept: $(0, -27)$

Part 1 Answer

The intercepts are:
(9, 0), (-3, 0), (0, -27)

Would you like a detailed explanation of how to test for symmetry for this equation, or do you have any other questions?

Here are 5 related questions to further explore this topic:

1. How do you factorize a quadratic equation to find the x-intercepts?
2. What does it mean for a graph to have symmetry with respect to the y-axis?
3. How can you determine the vertex of the parabola given by the equation $y = x^2 - 6x - 27$?
4. What is the axis of symmetry for the parabola represented by $y = x^2 - 6x - 27$?
5. How do you determine if a quadratic equation will open upwards or downwards?

Tip: When solving for intercepts, always remember to substitute the appropriate values (0 for x when finding y-intercepts and 0 for y when finding x-intercepts).