To graph the quadratic equation $y = x^2 - 2x - 15$, follow these steps:

Step 1: Identify key features

1. Determine the vertex:
   The equation is in standard form $y = ax^2 + bx + c$, where $a = 1$, $b = -2$, and $c = -15$.
   The x-coordinate of the vertex is found using the formula: $x = -\frac{b}{2a}$ Substituting values: $x = -\frac{-2}{2(1)} = 1$ To find the y-coordinate of the vertex, substitute $x = 1$ back into the equation: $y = (1)^2 - 2(1) - 15 = -16$ So, the vertex is $(1, -16)$.

2. Find the y-intercept:
   Substitute $x = 0$: $y = (0)^2 - 2(0) - 15 = -15$ The y-intercept is $(0, -15)$.

3. Find the x-intercepts (roots):
   Solve $x^2 - 2x - 15 = 0$ using factoring: $x^2 - 2x - 15 = (x - 5)(x + 3)$ Set each factor equal to 0: $x - 5 = 0 \quad \text{or} \quad x + 3 = 0$ $x = 5 \quad \text{and} \quad x = -3$ The x-intercepts are $(5, 0)$ and $(-3, 0)$.

Step 2: Sketch the graph

1. Plot the vertex $(1, -16)$, the y-intercept $(0, -15)$, and the x-intercepts $(-3, 0)$ and $(5, 0)$.
2. Draw a smooth parabola that opens upwards (since $a > 0$) and passes through these points.

Would you like me to create the graph for you? Alternatively, I can provide more details on transformations or graphing tips.

Related Questions

1. How do you convert a quadratic equation from standard form to vertex form?
2. What does the sign of $a$ tell you about the direction of a parabola?
3. How would you find the axis of symmetry for this graph?
4. How do the roots relate to the vertex in a symmetric parabola?
5. What are some practical applications of graphing quadratic equations?

Tip: Always check for symmetry about the vertex when graphing a parabola to ensure accuracy.