Question

which function displays this end behavior?

• as x approaches negative infinity, y approaches negative infinity.
• as x approaches positive infinity, y approaches negative infinity.

a. $y = -3x^2 + 4$
b. $y = (2x)^{\frac{1}{2}} - 7$
c. $y = -2x^3 - 1$
d. $y = -x^{\frac{1}{2}} + 5$

Explanation:

Step1: Analyze the end behavior of a quadratic function (Option A)

For a quadratic function \( y = ax^2 + bx + c \), the end behavior is determined by the leading coefficient \( a \) and the degree (2, even). If \( a < 0 \), as \( x \to \pm\infty \), \( y \to -\infty \). But let's check other options too.

Step2: Analyze the end behavior of a square - root function (Option B)

The function \( y=(2x)^{\frac{1}{2}}-7=\sqrt{2x}-7 \) has a domain \( x\geq0 \). So we can't consider the behavior as \( x\to-\infty \) since the function is not defined for \( x < 0 \). So this option is out.

Step3: Analyze the end behavior of a cubic function (Option C)

For a cubic function \( y = ax^3+bx^2 + cx + d \), the end behavior is determined by the leading coefficient \( a \) and the degree (3, odd). If \( a<0 \), as \( x\to+\infty \), \( y\to-\infty \) and as \( x\to-\infty \), \( y\to+\infty \) (because for odd degree, as \( x\to-\infty \), \( x^3\to-\infty \), and multiplying by a negative \( a \) gives \( y\to+\infty \)). So this does not match the given end behavior.

Step4: Analyze the end behavior of a square - root function (Option D)

The function \( y=-x^{\frac{1}{2}} + 5=-\sqrt{x}+5 \) has a domain \( x\geq0 \). So we can't consider the behavior as \( x\to-\infty \) since the function is not defined for \( x < 0 \). So this option is out. Wait, but earlier analysis of Option A: For \( y=-3x^2 + 4 \), degree \( n = 2 \) (even), leading coefficient \( a=-3<0 \). So as \( x\to+\infty \), \( x^2\to+\infty \), \( - 3x^2\to-\infty \), so \( y\to-\infty \). As \( x\to-\infty \), \( x^2\to+\infty \), \( -3x^2\to-\infty \), so \( y\to-\infty \). Which matches the given end behavior. Wait, but earlier analysis of Option C was wrong? Wait no, let's re - analyze Option C: \( y=-2x^3-1 \), degree \( n = 3 \) (odd), leading coefficient \( a=-2<0 \). As \( x\to+\infty \), \( x^3\to+\infty \), \( -2x^3\to-\infty \), so \( y\to-\infty \). As \( x\to-\infty \), \( x^3\to-\infty \), \( -2x^3\to+\infty \), so \( y\to+\infty \). So it doesn't match. Option A: quadratic, \( a=-3<0 \), degree 2 (even). So as \( x\to\pm\infty \), \( y\to-\infty \), which matches the given end behavior.

Answer:

A. \( y=-3x^{2}+4 \)