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On what interval is the square root function \( y=\sqrt{2 x+6}+4 \) increasing? (1 point) The function is increasing over the interval \( [\square, \infty) \). Check answer Remaining Attempts : 3

Ask by Klein Mathis. in the United States
Jan 21,2025

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Answer

The function \( y = \sqrt{2x + 6} + 4 \) is increasing on the interval \([-3, \infty)\).

Solution

The function \( y = \sqrt{2x + 6} + 4 \) is defined when the expression inside the square root is non-negative: \[ 2x + 6 \geq 0 \implies x \geq -3 \] Since the square root function is increasing for all \( x \) in its domain, the function \( y = \sqrt{2x + 6} + 4 \) is increasing on the interval: \[ [-3, \infty) \] **Answer:** \([-3, \infty)\)

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Bonus Knowledge

To find where the function \( y = \sqrt{2x + 6} + 4 \) is increasing, we can analyze the expression inside the square root. The function is defined for \( 2x + 6 \geq 0 \), which simplifies to \( x \geq -3 \). Since the square root function always has a positive derivative when we are in its domain, and since adding 4 is a constant shift upwards, we know that the function is indeed increasing for all \( x \) in the interval \( [-3, \infty) \). So, the function is increasing over the interval \( [-3, \infty) \). To check common mistakes, many students might overlook the domain of the square root function. Always ensure you define the domain before analyzing the intervals of increase or decrease. If you forget to check for where the function is defined, you could misidentify the intervals!

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