Graphing polynomial functions

An SAT Math micro-topic under Polynomial and other nonlinear graphs (Advanced Math). Free to read — no account needed.

The zeros of a polynomial are the xx-values where its graph touches or crosses the xx-axis.
How the graph behaves at a zero depends on whether that zero is single, double, or higher.
double_zero.png
At a single zero, the graph passes straight through the axis.
It comes from one factor like (x+2)(x + 2), and the curve is on one side of the axis before the zero and the other side after it.
This is the ordinary "crossing" behaviour.
At a double zero, the graph touches the axis and turns back without crossing.
It comes from a squared factor like (x−1)2(x - 1)^2, so the curve dips down to the axis and bounces back up (or peaks up and comes back down).
Changing direction right at the zero is the signature of a double zero.
So to read a polynomial graph, note where it meets the axis and how it behaves there.
A clean crossing signals a single zero; a touch-and-turn signals a double zero.
This lets you match a graph to its factored form, since each squared factor produces a bounce.

Worked examples

The graph below is a parabola.
polygraph_ex1.png
Which polynomial does it represent?
(A) (x+1)(x−2)(x + 1)(x - 2)
(B) (x−1)(x+2)(x - 1)(x + 2)
The graph crosses the xx-axis at x=−1x = -1 and x=2x = 2, so its factors are (x+1)(x + 1) and (x−2)(x - 2).
Choice (A) has exactly these factors, so (A) is correct.
The graph below is a cubic.
polygraph_ex2.png
Which polynomial does it represent?
(A) (x+3)(x−2)2(x + 3)(x - 2)^2
(B) (x+3)(x−2)(x + 3)(x - 2)
The graph crosses at x=−3x = -3 but only touches and turns at x=2x = 2, so x=2x = 2 is a double zero that needs a squared factor.
Choice (A) has (x−2)2(x - 2)^2, so (A) is correct.
The graph below is a downward parabola.
polygraph_ex3.png
Which polynomial does it represent?
(A) −x(x−3)-x(x - 3)
(B) x(x−3)x(x - 3)
The graph crosses at x=0x = 0 and x=3x = 3 and opens downward, so the leading coefficient must be negative.
Choice (A) carries the negative sign, so (A) is correct.

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