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.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

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.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations