Question

question 7 of 10
what can you say about the y-values of the two functions $f(x) = 3x^2 - 3$ and $g(x) = 2^x - 3$?

a. $f(x)$ and $g(x)$ have equivalent minimum y-values.

b. the minimum y-value of $g(x)$ approaches -3.

c. $f(x)$ has the smallest possible y-value.

d. $g(x)$ has the smallest possible y-value.

Explanation:

Step1: Analyze \( f(x) = 3x^2 - 3 \)

This is a quadratic function in the form \( ax^2 + bx + c \) with \( a = 3>0 \), so it opens upward. The vertex (minimum point) occurs at \( x = -\frac{b}{2a} \). Here, \( b = 0 \), so \( x = 0 \). Substituting \( x = 0 \) into \( f(x) \), we get \( f(0)=3(0)^2 - 3=-3 \). So the minimum \( y \)-value of \( f(x) \) is \( -3 \).

Step2: Analyze \( g(x) = 2^x - 3 \)

The function \( y = 2^x \) is an exponential function with base \( 2>1 \), so it is an increasing function. The range of \( 2^x \) is \( (0, +\infty) \) because as \( x ightarrow -\infty \), \( 2^x ightarrow 0 \), and as \( x ightarrow +\infty \), \( 2^x ightarrow +\infty \). For \( g(x)=2^x - 3 \), we subtract 3 from \( 2^x \). So the range of \( g(x) \) is \( (-3, +\infty) \). This means as \( x ightarrow -\infty \), \( g(x) ightarrow -3 \), but it never actually reaches \( -3 \); the minimum \( y \)-value approaches \( -3 \).

Step3: Evaluate the options

- Option A: \( f(x) \) has a minimum \( y \)-value of \( -3 \) (attained), while \( g(x) \) approaches \( -3 \) (not attained). So they are not equivalent. Eliminate A.
- Option B: From the analysis of \( g(x) \), the minimum \( y \)-value approaches \( -3 \). This is correct.
- Option C: \( f(x) \) has a minimum \( y \)-value of \( -3 \), but \( g(x) \) can get arbitrarily close to \( -3 \) (but not less than or equal to \( -3 \) in the way \( f(x) \) does). Wait, no—\( f(x) \) has a minimum of \( -3 \), and \( g(x) \) approaches \( -3 \) from above (since its range is \( (-3, +\infty) \)). So \( f(x) \) has a minimum \( y \)-value, but \( g(x) \) does not have a smallest possible \( y \)-value (it can get as close to \( -3 \) as we want but never reaches it or goes below). Wait, re - evaluating: The question is about "smallest possible \( y \)-value". \( f(x) \) has a smallest (minimum) \( y \)-value of \( -3 \). \( g(x) \) does not have a smallest \( y \)-value because it can get closer and closer to \( -3 \) but never actually reaches a smallest value (since it's an open interval \( (-3, +\infty) \)). But let's check the options again. Option B says the minimum \( y \)-value of \( g(x) \) approaches \( -3 \), which is correct. Option C: Wait, maybe I made a mistake earlier. Let's re - check. \( f(x) \) has a minimum \( y \)-value of \( -3 \). \( g(x) \) has a range of \( (-3, +\infty) \), so the \( y \)-values of \( g(x) \) are greater than \( -3 \), and \( f(x) \) has a \( y \)-value of \( -3 \). So \( f(x) \) has a smaller \( y \)-value (since \( -3 \) is less than any value in \( (-3, +\infty) \))? Wait, no—if \( g(x) \) is in \( (-3, +\infty) \), then all \( y \)-values of \( g(x) \) are greater than \( -3 \), and \( f(x) \) has a \( y \)-value of \( -3 \). So \( f(x) \) has a smaller \( y \)-value? But option C says " \( f(x) \) has the smallest possible \( y \)-value". Wait, but \( g(x) \) does not have a smallest possible \( y \)-value (because it can get infinitely close to \( -3 \) but never reaches a minimum). \( f(x) \) does have a smallest (minimum) \( y \)-value of \( -3 \). But let's go back to the options. Option B is correct about \( g(x) \)'s minimum approaching \( -3 \). Option D: \( g(x) \) does not have a smallest possible \( y \)-value (since it can get as close to \( -3 \) as we want but never reaches it or goes below), so D is wrong. Option C: Wait, \( f(x) \) has a minimum \( y \)-value of \( -3 \), and \( g(x) \)'s \( y \)-values are all greater than \( -3 \). So \( f(x) \) has a smaller \( y \)-value (the minimum is \( -3 \)), but does it have the "smallest possible \( y \)-value"? Since \( g(x) \) can't go below \( -3 \) (its range is \( (-3, +\infty) \)), \( f(x) \) has a \( y \)-value of \( -3 \), which is smaller than any \( y \)-value of \( g(x) \). But the key is the analysis of each function. However, the correct option from the analysis of the minimum of \( g(x) \) is B. Wait, maybe my initial analysis of C was wrong. Let's re - check the options:

- Option B: The minimum \( y \)-value of \( g(x) \) approaches \( -3 \). This is correct because as \( x ightarrow -\infty \), \( g(x)=2^x - 3 ightarrow 0 - 3=-3 \).
- Option C: \( f(x) \) has a minimum \( y \)-value of \( -3 \), but \( g(x) \) does not have a "smallest possible \( y \)-value" in the sense of attaining a minimum. Wait, the term "smallest possible \( y \)-value"—for \( f(x) \), it's \( -3 \) (attained). For \( g(x) \), since it can get arbitrarily close to \( -3 \) but never reaches it, it does not have a smallest possible \( y \)-value (because there's no single smallest value; you can always find a \( y \)-value closer to \( -3 \) than a previous one). So \( f(x) \) has a smallest possible \( y \)-value (the minimum), and \( g(x) \) does not. But the option C says " \( f(x) \) has the smallest possible \( y \)-value"—but is that true? Let's see the \( y \)-values: \( f(x) \) can be \( -3 \), and \( g(x) \) is always greater than \( -3 \). So \( f(x) \)'s minimum is \( -3 \), and \( g(x) \)'s \( y \)-values are all greater than \( -3 \). So \( f(x) \) has a smaller \( y \)-value (the minimum is \( -3 \)), but does it have the "smallest possible \( y \)-value" among the two? Wait, the question is about the \( y \)-values of the two functions. Let's re - evaluate the options with the correct analysis:

1. \( f(x)=3x^2 - 3 \): quadratic, opens up, vertex at \( (0, -3) \), so minimum \( y=-3 \) (attained).
2. \( g(x)=2^x - 3 \): exponential, \( 2^x>0 \) for all \( x \), so \( g(x)> - 3 \) for all \( x \), and as \( x ightarrow -\infty \), \( g(x) ightarrow -3 \).

Now:

- Option A: \( f(x) \) attains \( -3 \), \( g(x) \) approaches \( -3 \) (never attains it), so their minimum \( y \)-values are not equivalent. A is wrong.
- Option B: As \( x ightarrow -\infty \), \( g(x) ightarrow -3 \), so the minimum \( y \)-value (the infimum) approaches \( -3 \). B is correct.
- Option C: \( f(x) \) has a minimum \( y \)-value of \( -3 \), and \( g(x) \)'s \( y \)-values are all greater than \( -3 \). So \( f(x) \) has a \( y \)-value ( \( -3 \)) that is smaller than any \( y \)-value of \( g(x) \). But does \( f(x) \) have the "smallest possible \( y \)-value"? Since \( g(x) \) can't go below \( -3 \), \( f(x) \)'s minimum of \( -3 \) is the smallest \( y \)-value between the two functions? Wait, but the question is about the \( y \)-values of each function. For \( f(x) \), the smallest possible \( y \)-value is \( -3 \) (it's a minimum). For \( g(x) \), there is no smallest possible \( y \)-value (because it can get closer and closer to \( -3 \) without bound). So in terms of the functions' own \( y \)-values: \( f(x) \) has a smallest (minimum) \( y \)-value, \( g(x) \) does not. But the option C says " \( f(x) \) has the smallest possible \( y \)-value"—but is that in comparison or for its own function? The question is "What can you say about the \( y \)-values of the two functions". Let's re - read the options. Option B is correct about \( g(x) \)'s minimum approaching \( -3 \). Option C: If we consider the two functions, \( f(x) \) has a \( y \)-value of \( -3 \), and \( g(x) \) is always above \( -3 \), so \( f(x) \) has a smaller \( y \)-value, but does it have the "smallest possible \( y \)-value"? Wait, maybe the key is that \( f(x) \) has a minimum (a smallest \( y \)-value), while \( g(x) \) does not have a smallest \( y \)-value (since it's an open interval). But the option B is correct because it accurately describes \( g(x) \)'s behavior.

Answer:

B. The minimum \( y \)-value of \( g(x) \) approaches \( -3 \)