You can sketch a polynomial's graph without plotting dozens of points. Combine three facts: the End Behavior of Polynomials, the zeros and their multiplicities, and the $y$-intercept.
Key idea
To sketch $f$: (1) find the zeros and their multiplicities from the factored form; (2) find the end behavior from the leading term; (3) find $f(0)$; (4) draw from left to right, crossing or touching at each zero. A degree-$n$ graph has at most $n - 1$ turning points.
Why it works
A polynomial graph is one smooth, unbroken curve. It can only change sign at a zero, and it changes sign there only when the multiplicity is odd. Once you know where it is positive and negative and how it behaves at the ends, the rough shape is forced. Between zeros the graph turns around somewhere, but finding exactly where needs calculus.
Worked examples
Worked example: A full sketch plan
Sketch $f(x) = (x + 2)(x - 1)^2(x - 3)$.
- Zeros: $-2$ (multiplicity 1, crosses), $1$ (multiplicity 2, touches), $3$ (multiplicity 1, crosses).
- Leading term: $x \cdot x^2 \cdot x = x^4$. Even degree, positive, so both ends rise.
- $y$-intercept: $f(0) = (2)(-1)^2(-3) = -6$.
- Signs: left of $-2$ the graph is above the axis (the left end rises). It crosses at $-2$ and stays below until $1$, where it touches the axis and turns back down. It stays below until it crosses at $3$, then rises.
- Check one interval with a number: $f(2) = (4)(1)(-1) = -4 < 0$, so the graph is below the axis between 1 and 3, as planned.
Answer: From the upper left, cross at $-2$, dip below (through $(0, -6)$), touch at $1$, dip again, cross at $3$, then up to the right.
Worked example: Counting turning points
At most how many turning points can $g(x) = -x^5 + 4x^3$ have?
- The degree is 5, so at most $5 - 1 = 4$.
- Factor to see the zeros: $-x^3(x^2 - 4) = -x^3(x - 2)(x + 2)$, zeros $-2, 0, 2$.
Answer: At most 4 (this one has exactly 2).
Common mistake
Making every zero a crossing point, or ignoring the end behavior. Check the sign in each interval with one test number; the picture must agree with all of them.