Construct polynomial and rational graphs

Try

A formula can tell you where a graph crosses an axis, where it touches without crossing, and where no point exists. Start with P(x)=(x+1)(x−2)². Its roots are −1 and 2, but their factors have different powers. Predict whether the graph changes sign at both roots. Do not join a few plotted points until you have checked those signs.

For the rational expression R(x)=(x²−1)/(x²−x−2), first find where its denominator is zero. A simplified formula must retain any exclusions from the original expression.

Predict

The squared factor in P is positive away from 2, so its sign is the sign of x+1. Predict a crossing at −1 and a touch at 2. In R, factoring gives (x−1)(x+1)/[(x−2)(x+1)]. Canceling x+1 is legal only when x≠−1. Predict a missing point at −1 and different behavior at 2, where the remaining denominator still vanishes.

Experiment

Input	P(x)	R(x)
−1	0	Undefined; hole height would be 2/3
0	4	1/2
1	2	0
2	0	Undefined; vertical asymptote
3	4	2

Read the two zeros of P from the graph, then use a sign check on either side to distinguish crossing from touching. For R, explain why the open circle is not an intercept even though the reduced formula has a value there. Its vertical location is a limiting height, not an attained output at that input.

Understand

A polynomial factor (x−r)ᵐ gives a root r of multiplicity m when the other factors do not vanish there. For odd m, that factor reverses sign across r; for even m, it does not. The nonzero remaining factors have a fixed sign sufficiently near r, so the full graph respectively crosses or stays on the same side. Multiplicity describes this local behavior; it does not by itself locate every turning point.

Expanding P gives x³−3x²+4. Its leading term controls its ends: P(x)/x³=1−3/x+4/x³ approaches 1 for large positive or negative x. Therefore the left end tends downward without bound and the right end upward. The exact landmarks and sign intervals constrain the sketch; the supplied curve shows their connecting behavior. A finite window does not impose finite domain endpoints.

For R, the original denominator is zero at −1 and 2. On the remaining domain, R(x)=(x−1)/(x−2)=1+1/(x−2). At −1 the reduced expression has the finite value 2/3, but the original input remains excluded: this is a removable hole. Near 2, the reciprocal term grows negatively without bound from the left and positively without bound from the right, giving a vertical asymptote. As x goes far in either direction, that term tends to zero, so y=1 is a horizontal asymptote.

An x-intercept needs both a zero output and a valid input. Thus R has its x-intercept at 1, not −1. Its y-intercept is (0,1/2). To sketch a rational graph, record domain exclusions before canceling, then use the reduced expression for values, signs and end behavior. Mark holes with open circles and never draw a vertical connector between separate branches.

Apply

Construct a new graph from the practice formula rather than copying the supplied curve. Include root multiplicities, intercepts, domain, sign intervals and end behavior; for a rational expression also mark holes and asymptotes. An exact coordinate-and-branch description is an accessible alternative to a drawing, but a list of isolated values alone does not describe the branches.

Continue

These constructions address polynomial and rational representation actions in F-IF.C.7c/d. They do not make a plotted sketch a proof of every feature: use factorization, sign reasoning and limiting comparisons to justify what the picture claims. Next, interpret quadratic graphs and combined functions in a bounded context.

Continue to Combine meaningful functions into a quadratic decision.