Question

for the following exercises, determine whether the relation represents ( y ) as a function of ( x ).

1. ( 5x + 2y = 10 )
2. ( y = x^2 )
3. ( x = y^2 )
4. ( 3x^2 + y = 14 )
5. ( 2x + y^2 = 6 )
6. ( y = -2x^2 + 40x )
7. ( y = \frac{1}{x} )
8. ( x = \frac{3y + 5}{7y - 1} )
9. ( x = sqrt{1 - y^2} )
10. ( y = \frac{3x + 5}{7x - 1} )
11. ( x^2 + y^2 = 9 )
12. ( 2xy = 1 )
13. ( x = y^3 )
14. ( y = x^3 )
15. ( y = sqrt{1 - x^2} )
16. ( x = pm sqrt{1 - y} )
17. ( y = pm sqrt{1 - x} )
18. ( y^2 = x^2 )

Answer:

To determine if \( y \) is a function of \( x \), we use the vertical line test: a relation represents \( y \) as a function of \( x \) if every \( x \)-value has at most one \( y \)-value. We analyze each problem:

Problem 8: \( 5x + 2y = 10 \)

Step 1: Solve for \( y \)

Rearrange: \( 2y = -5x + 10 \) → \( y = -\frac{5}{2}x + 5 \).

Step 2: Apply Vertical Line Test

For any \( x \), there is exactly one \( y \) (linear equation, slope \( -\frac{5}{2} \)).
Conclusion: \( y \) is a function of \( x \).

Problem 9: \( y = x^2 \)

Step 1: Analyze the Relation

For any \( x \), squaring \( x \) gives exactly one \( y \) (e.g., \( x=2 \) → \( y=4 \); \( x=-2 \) → \( y=4 \), but each \( x \) has one \( y \)).

Step 2: Vertical Line Test

A vertical line intersects the parabola at most once.
Conclusion: \( y \) is a function of \( x \).

Problem 10: \( x = y^2 \)

Step 1: Solve for \( y \)

\( y = \pm\sqrt{x} \) (e.g., \( x=4 \) → \( y=2 \) or \( y=-2 \)).

Step 2: Vertical Line Test

A vertical line (e.g., \( x=4 \)) intersects the graph at two points (\( (4,2) \) and \( (4,-2) \)).
Conclusion: \( y \) is not a function of \( x \).

Problem 11: \( 3x^2 + y = 14 \)

Step 1: Solve for \( y \)

\( y = -3x^2 + 14 \).

Step 2: Vertical Line Test

For any \( x \), there is exactly one \( y \) (quadratic, opens downward).
Conclusion: \( y \) is a function of \( x \).

Problem 12: \( 2x + y^2 = 6 \)

Step 1: Solve for \( y \)

\( y^2 = -2x + 6 \) → \( y = \pm\sqrt{-2x + 6} \) (e.g., \( x=1 \) → \( y=\pm2 \)).

Step 2: Vertical Line Test

A vertical line (e.g., \( x=1 \)) intersects the graph at two points.
Conclusion: \( y \) is not a function of \( x \).

Problem 13: \( y = -2x^2 + 40x \)

Step 1: Analyze the Relation

Quadratic function: for any \( x \), substituting \( x \) gives exactly one \( y \).

Step 2: Vertical Line Test

A vertical line intersects the parabola at most once.
Conclusion: \( y \) is a function of \( x \).

Problem 14: \( y = \frac{1}{x} \)

Step 1: Analyze the Relation

For \( x
eq 0 \), \( \frac{1}{x} \) gives exactly one \( y \) (e.g., \( x=2 \) → \( y=\frac{1}{2} \); \( x=-2 \) → \( y=-\frac{1}{2} \)).

Step 2: Vertical Line Test

No vertical line intersects the hyperbola more than once (excluding \( x=0 \), which is undefined).
Conclusion: \( y \) is a function of \( x \).

Problem 15: \( x = \frac{3y + 5}{7y - 1} \)

Step 1: Solve for \( y \) (optional, but analyze \( x \)-to-\( y \))

This is a rational function in \( y \). For a given \( x \), solving \( x(7y - 1) = 3y + 5 \) gives a linear equation in \( y \), so at most one \( y \) (except when \( 7x - 3 = 0 \), but \( x \) is defined such that \( 7y - 1
eq 0 \)).

Step 2: Vertical Line Test

Each \( x \) (in domain) maps to one \( y \).
Conclusion: \( y \) is a function of \( x \).

Problem 16: \( x = \sqrt{1 - y^2} \)

Step 1: Solve for \( y \)

Square both sides: \( x^2 = 1 - y^2 \) → \( y^2 = 1 - x^2 \) → \( y = \pm\sqrt{1 - x^2} \) (e.g., \( x=0 \) → \( y=\pm1 \)).

Step 2: Vertical Line Test

A vertical line (e.g., \( x=0 \)) intersects the circle at two points.
Conclusion: \( y \) is not a function of \( x \).

Problem 17: \( y = \frac{3x + 5}{7x - 1} \)

Step 1: Analyze the Relation

Rational function: for \( x
eq \frac{1}{7} \), substituting \( x \) gives exactly one \( y \).

Step 2: Vertical Line Test

No vertical line intersects the graph more than once (excluding \( x=\frac{1}{7} \), undefined).
Conclusion: \( y \) is a function of \( x \).

Problem 18: \( x^2 + y^2 = 9 \)

Step 1: Solve for \( y \)

\( y^2 = 9 - x^2 \) → \( y = \pm\sqrt{9 - x^2} \) (e.g., \( x=0 \) → \( y=\pm3 \)).

Step 2: Vertical Line Test

A vertical line (e.g., \( x=0 \)) intersects the circle at two points.
Conclusion: \( y \) is not a function of \( x \).

Problem 19: \( 2xy = 1 \)

Step 1: Solve for \( y \)

\( y = \frac{1}{2x} \) (for \( x
eq 0 \)).

Step 2: Vertical Line Test

For \( x
eq 0 \), each \( x \) maps to one \( y \) (hyperbola).
Conclusion: \( y \) is a function of \( x \).

Problem 20: \( x = y^3 \)

Step 1: Solve for \( y \)

\( y = \sqrt[3]{x} \) (cube root, which is one-to-one: \( x=8 \) → \( y=2 \); \( x=-8 \) → \( y=-2 \)).

Step 2: Vertical Line Test

Each \( x \) maps to exactly one \( y \).
Conclusion: \( y \) is a function of \( x \).

Problem 21: \( y = x^3 \)

Step 1: Analyze the Relation

Cubic function: for any \( x \), cubing \( x \) gives exactly one \( y \) (one-to-one, since \( y' = 3x^2 \geq 0 \)).

Step 2: Vertical Line Test

A vertical line intersects the cubic at most once.
Conclusion: \( y \) is a function of \( x \).

Problem 22: \( y = \sqrt{1 - x^2} \)

Step 1: Analyze the Relation

Square root function: for \( x \in [-1, 1] \), \( \sqrt{1 - x^2} \) gives exactly one non-negative \( y \) (e.g., \( x=0 \) → \( y=1 \); \( x=1 \) → \( y=0 \)).

Step 2: Vertical Line Test

A vertical line intersects the upper half of the circle at most once.
Conclusion: \( y \) is a function of \( x \).

Problem 23: \( x = \pm\sqrt{1 - y} \)

Step 1: Solve for \( y \)

Square both sides: \( x^2 = 1 - y \) → \( y = 1 - x^2 \). Wait, no—original is \( x = \pm\sqrt{1 - y} \), so \( 1 - y \geq 0 \) → \( y \leq 1 \), and \( x \) can be positive or negative for the same \( y \). Wait, the relation is \( x = \pm\sqrt{1 - y} \), so for a given \( y \), \( x \) has two values (e.g., \( y=0 \) → \( x=\pm1 \)). But we check if \( y \) is a function of \( x \): solve for \( y \): \( x^2 = 1 - y \) → \( y = 1 - x^2 \).

Step 2: Vertical Line Test

For any \( x \), \( y = 1 - x^2 \) gives exactly one \( y \) (quadratic, opens downward).
Conclusion: \( y \) is a function of \( x \).

Problem 24: \( y = \pm\sqrt{1 - x} \)

Step 1: Analyze the Relation

For \( x \leq 1 \), \( y \) has two values (e.g., \( x=0 \) → \( y=\pm1 \)).

Step 2: Vertical Line Test

A vertical line (e.g., \( x=0 \)) intersects the graph at two points.
Conclusion: \( y \) is not a function of \( x \).

Problem 25: \( y^2 = x^2 \)

Step 1: Solve for \( y \)

\( y = \pm x \) (e.g., \( x=2 \) → \( y=\pm2 \)).

Step 2: Vertical Line Test

A vertical line (e.g., \( x=2 \)) intersects the graph at two points.
Conclusion: \( y \) is not a function of \( x \).

Summary of Results:

Problem	\( y \) is a function of \( x \)?
9	Yes
10	No
11	Yes
12	No
13	Yes
14	Yes
15	Yes
16	No
17	Yes
18	No
19	Yes
20	Yes
21	Yes
22	Yes
23	Yes
24	No
25	No

(Note: For clarity, recheck Problem 23: original is \( x = \pm\sqrt{1 - y} \), but solving for \( y \) gives \( y = 1 - x^2 \), which is a function. Problem 24: \( y = \pm\sqrt{1 - x} \) is not a function of \( x \) because one \( x \) gives two \( y \)-values.)