Question

question 7 (1 point)

which relation will pass the vertical line test?

a) (y = 4 pm sqrt{x})

b) (x^2 + y^2 = 25)

c) (3x^2 + 5y^2 = 18)

d) (y = 6 - sqrt{2x + 4})

Explanation:

Understand the vertical line test

The vertical line test determines if a relation is a function. A relation is a function if and only if every vertical line intersects its graph at most once. Algebraically, this means that for each input \(x\) in the domain, there is exactly one corresponding output \(y\).

Analyze option A

The equation is:

$$y = 4 \pm \sqrt{x}$$

For any \(x > 0\), there are two distinct values of \(y\). For example, if \(x = 1\):

$$y = 4 + 1 = 5 \quad \text{or} \quad y = 4 - 1 = 3$$

Since one \(x\)-value maps to two \(y\)-values, a vertical line at \(x = 1\) intersects the graph twice. This relation fails the test.

Analyze option B

The equation is:

$$x^2 + y^2 = 25$$

This represents a circle centered at the origin with radius 5. Solving for \(y\):

$$y = \pm\sqrt{25 - x^2}$$

For any \(x\) in the interval \((-5, 5)\), there are two distinct \(y\)-values. For example, if \(x = 0\), then \(y = 5\) or \(y = -5\). Thus, it fails the test.

Analyze option C

The equation is:

$$3x^2 + 5y^2 = 18$$

This represents an ellipse. Solving for \(y\):

$$5y^2 = 18 - 3x^2 \implies y = \pm\sqrt{\frac{18 - 3x^2}{5}}$$

For values of \(x\) where the radicand is positive, there are two distinct \(y\)-values for each \(x\). Thus, it fails the test.

Analyze option D

The equation is:

$$y = 6 - \sqrt{2x + 4}$$

For any \(x \ge -2\), the principal square root yields a single, unique real value. Therefore, for each input \(x\) in the domain, there is exactly one output \(y\). A vertical line will intersect this graph at most once, so it passes the test.

Answer:

• (A) \(y = 4 \pm \sqrt{x}\)
• (B) \(x^2 + y^2 = 25\)
• (C) \(3x^2 + 5y^2 = 18\)
• (D) \(y = 6 - \sqrt{2x + 4}\) (Correct answer)