7.4 Graphs, Domain, and Range of Radical and Rational Functions

7.4 Graphs, Domain, and Range of Radical and Rational Functions

Table 11’s remaining radical/rational skills are domain and range and graphing. The algebra of the last three sections tells you which x-values are legal and which candidates were extra. The graph makes those restrictions visible: a square-root curve that starts at an endpoint, or a rational graph that never crosses a vertical line.

Function-notation domain from Function Notation, Evaluation, Domain, and Range is the same list you graph here: polynomials all reals, √(g(x)) needs g(x) ≥ 0, and p(x)/q(x) needs q(x) ≠ 0. Transformations from Transformations and Interpreting Functions in Context move those graphs without changing the checklist. Work FREE graph-reading items at /practice/accuplacer-advanced-algebra.

Parent square-root graph y = √x

y = √x is defined for x ≥ 0. It starts at the origin (0, 0) and increases slowly through (1, 1), (4, 2), (9, 3), and (16, 4). The range is y ≥ 0. It is a function (vertical line test from the functions chapter), and it is the right half of the sideways parabola x = y^2 after you throw away y < 0.

The graph is not a straight line. Equal steps in y require growing steps in x. From y = 1 to y = 2 you move 3 horizontal units; from y = 2 to y = 3 you move 5. That flattening is how you distinguish y = √x from y = x and from y = x^2 on a four-graph item.

Shifts of square-root graphs

y = √(x − h) + k starts at (h, k).

• Inside: x − h ≥ 0 ⇒ domain x ≥ h.
• Outputs are k and up (the parent range shifted by k) ⇒ range y ≥ k when the graph is not reflected.

Worked: y = √(x + 3) − 2

This is y = √(x − (−3)) + (−2). The endpoint is (−3, −2). Domain: x + 3 ≥ 0 ⇒ x ≥ −3. Range: y ≥ −2. A table that confirms the shift: at x = −3, y = −2; at x = −2, y = √1 − 2 = −1; at x = 1, y = √4 − 2 = 0.

A common trap uses the “inside is opposite” rule on the vertical shift too, and claims the graph starts at (3, 2) or (−3, 2). Only the horizontal piece flips its sign inside the formula: x + 3 is a shift left 3, while − 2 outside is a shift down 2.

Further transformations:

• y = −√x reflects over the x-axis: domain still x ≥ 0, range y ≤ 0.
• y = √(−x) reflects over the y-axis: domain x ≤ 0, range y ≥ 0.
• y = 2√x stretches vertically: through (1, 2) instead of (1, 1); domain and range still [0, ∞).
• y = √(x − 4) + 1 starts at (4, 1), domain x ≥ 4, range y ≥ 1.

Cube-root parent y = ∛x has domain all reals and range all reals. It passes through (−8, −2), (0, 0), (8, 2) and does not stop at the origin. That is why ∛(x − 1) still has domain all reals, matching the solving section’s cube-root note.

Rational parent pieces: y = 1/x and y = (ax + b)/(cx + d)

The simplest rational graph is y = 1/x: vertical asymptote x = 0, horizontal asymptote y = 0, domain all reals except 0, range all reals except 0. Two branches, one in quadrant I and one in quadrant III. As x → 0+, y → +∞; as x → 0−, y → −∞; as |x| → ∞, y → 0.

A linear-over-linear function

y = (ax + b)/(cx + d) (with c ≠ 0)

is a shifted, scaled version of 1/x.

Worked: y = (2x + 1)/(x − 3)

Denominator zero at x = 3. Numerator at x = 3 is 7 ≠ 0, so this is a vertical asymptote, not a hole: VA x = 3. Equivalently, −d/c with c = 1, d = −3 is x = 3.

Degrees of numerator and denominator are both 1, leading coefficients 2 and 1, so HA y = 2.

x-intercept: 2x + 1 = 0 ⇒ x = −1/2 (allowed, since −1/2 ≠ 3).

y-intercept: f(0) = 1/(−3) = −1/3.

Domain: all reals except x = 3. Range: all reals except y = 2 — a non-constant linear-over-linear that does not reduce to a constant never attains its horizontal asymptote.

A quick sign check near the VA: just right of 3, say x = 4, y = 9/1 = 9 (large positive). Just left, x = 2, y = 5/(−1) = −5 (negative). The two branches jump from −∞ to +∞ across x = 3.

Hole versus asymptote

y = (x^2 − 4)/(x − 2) = [(x − 2)(x + 2)]/(x − 2) = x + 2 for x ≠ 2.

The simplified graph is the line y = x + 2 with a hole at (2, 4). There is no vertical asymptote at x = 2, because the factor canceled. Domain is still x ≠ 2. An AAF item that asks for the graph will show an open circle at (2, 4), not a dashed vertical line. The range is all reals except 4, because that output would have required the missing input x = 2.

If you forget to cancel, you might report VA x = 2. That is the discriminant of this skill: factor first, then classify each zero of the denominator as a hole (canceled) or a VA (survived). The same discipline was in Rational Expressions and Equations: canceling (x − 3) from (x^2 − 9)/(x^2 − x − 6) produced a hole at x = 3 and a surviving exclusion x = −2 that is a VA of the simplified formula (x + 3)/(x + 2).

Horizontal asymptotes from degrees

Let N be the degree of the numerator and D the degree of the denominator.

AAF linear-over-linear items are the N = D case. y = (5x − 1)/(x + 4) has HA y = 5 and VA x = −4. y = 7/(x − 2) has N < D, so HA y = 0 and VA x = 2. y = (x^2 + 1)/(x − 1) has N = D + 1, so a slant asymptote (divide to get y = x + 1 plus a remainder); AAF may still ask the VA x = 1 and the domain x ≠ 1 without naming the slant line.

Domain and range, collected

Do not confuse domain with “where the graph is positive.” y = √x − 4 has domain x ≥ 0 even though outputs are negative until x = 16. Domain is allowed inputs; range is attained outputs.

Creating the function from a graph description

Table 11’s “creating … functions” is the translation from features to a formula. “A square-root graph starts at (−3, −2) and increases” is y = √(x + 3) − 2 (or a positive stretch of that, if a second point is given). “A rational graph never crosses x = 3 and flattens toward y = 2” matches any linear-over-linear with VA x = 3 and HA y = 2; one extra point locks the remaining coefficient. Then write the domain that goes with the formula: x ≠ 3, or x ≥ −3, not a vague “all x except the intercepts.”

Connecting the four skills in this chapter

Simplifying tells you the formula a graph actually follows after canceling. Solving tells you intercepts (set y = 0, or set two expressions equal). Domain exclusions from even roots and zero denominators are the same numbers you used to throw away extraneous roots. If a candidate solution of √(2x + 3) + 1 = 6 had been x = −2, it would also have been a point the graph of y = √(2x + 3) does not contain.

An early-band item may only ask you to evaluate or to read a starting point. Skills Insight 250–262 rewrites the rational expression; 263–275 adds two unlike denominators. Graphing and range reuse the functions chapter’s domain/range language, now specialized to radicals and rationals. Mixed FREE practice at /practice/accuplacer-advanced-algebra is how you find out which of those four Table 11 skills the CAT is currently emphasizing for you. Related placement tests in the same suite are ACCUPLACER Quantitative Reasoning, Algebra, and Statistics and ACCUPLACER Arithmetic; AAF is the higher algebra/functions placement, and this chapter is the radical/rational slice of that 20-item CAT.