One zero of $x^2-6x+k$ exceeds the other by 2. Find $k$.

Zeros of $p(x)$ are exactly the **$x$-coordinates where the graph $y=p(x)$ meets the $x$-axis**.

- **Linear** $y$ = ax + $b$: graph is a straight line, crosses $x$-axis ONCE → exactly 1 zero.
- **Quadratic** $y=a{x}^2+bx+c$: graph is a parabola. Possible cases:
- Crosses $x$-axis at 2 points → 2 distinct real zeros.
- Touches $x$-axis at 1 point → 1 repeated zero (vertex on $x$-axis).
- Does not meet $x$-axis → no real zeros.
- **Cubic** $y=a{x}^3$ + …: shape is a curve that crosses 1, 2, or 3 times → 1, 2, or 3 real zeros.

So 'find the zeros' is the algebraic version of 'find where the graph crosses the $x$-axis'.