Question

sketch the least positive angle θ and find the values of the six trigonometric functions of θ if the terminal side of an angle θ in standard position is defined by $-sqrt{10}x + y = 0, x ≥ 0$

which of the graphs shows $-sqrt{10}x + y = 0$, where $x ≥ 0$?
○ a. graph
○ b. graph
○ c. graph
○ d. graph

Explanation:

To determine which graph represents \( -\sqrt{10}x + y = 0 \) with \( x \geq 0 \), we first rewrite the equation in slope - intercept form (\( y=mx + b \), where \( m \) is the slope and \( b \) is the y - intercept).

Step 1: Rewrite the equation

Starting with \( -\sqrt{10}x + y=0 \), we can solve for \( y \) by adding \( \sqrt{10}x \) to both sides of the equation. We get \( y=\sqrt{10}x \).

The slope \( m = \sqrt{10}\approx3.16>0 \), and the y - intercept \( b = 0 \), which means the line passes through the origin \((0,0)\) and has a positive slope. Also, we have the restriction \( x\geq0 \), so we are only considering the part of the line \( y = \sqrt{10}x \) where the x - coordinate is non - negative.

Step 2: Analyze the slope and the domain

A positive slope means that as \( x \) increases (since \( x\geq0 \)), \( y \) also increases. Let's consider the general shape of the line \( y = mx \) with \( m>0 \) and \( x\geq0 \). The line starts at the origin and goes into the first quadrant (because when \( x>0 \), \( y=\sqrt{10}x>0 \)) with a steep positive slope (since \( \sqrt{10}\approx3.16 \) is a relatively large positive number).

Now let's analyze the options:

• Option A: If the line has a negative slope (going downwards from left to right) or is in the wrong quadrant, it can be eliminated. Since our line has a positive slope and \( x\geq0 \) (so we are in the first quadrant for \( x > 0\)), a line with a negative slope is incorrect.
• Option B: If the line is in the fourth quadrant (where \( y<0 \) when \( x>0 \)) or has a negative slope, it is incorrect. Our line \( y=\sqrt{10}x \) with \( x\geq0 \) has \( y\geq0 \) when \( x\geq0 \), so a line in the fourth quadrant is wrong.
• Option C: If the line is in the third quadrant (where \( x<0 \) and \( y<0 \)) or has a negative slope, it is incorrect. Our domain is \( x\geq0 \), so we do not consider the third quadrant.
• Option D: The line passes through the origin, has a positive slope, and since \( x\geq0 \), the part of the line we consider is in the first quadrant (for \( x>0 \)) with a steep positive slope, which matches our equation \( y = \sqrt{10}x,x\geq0 \).

Answer:

D