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Sagot :
Let's analyze the two functions \( f(x) = -5^x + 2 \) and \( g(x) = -5x^2 + 2 \) to determine their [tex]$y$[/tex]-values.
1. Function \( f(x) = -5^x + 2 \):
- This is an exponential function where the base of the exponent is 5 and it is negative.
- As \( x \) increases, \( 5^x \) grows exponentially. Since this term is negative, \( -5^x \) also grows negatively.
- Therefore, as \( x \) approaches positive infinity, \( -5^x \) approaches negative infinity.
- Thus, \( f(x) \) will get closer and closer to \( 2 \) from below (since we are adding 2 to a very large negative number).
[tex]\[ \lim_{{x \to \infty}} f(x) = 2 \][/tex]
However, \( f(x) \) never actually reaches 2; it only approaches it.
2. Function \( g(x) = -5x^2 + 2 \):
- This is a quadratic function opening downwards because the coefficient of \( x^2 \) is negative.
- The vertex of this parabola represents the maximum point since the parabola opens downwards.
- The vertex of a parabola \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \).
- In this case, \( a = -5 \), \( b = 0 \), and \( c = 2 \), so the vertex is at \( x = 0 \).
[tex]\[ g(0) = -5(0)^2 + 2 = 2 \][/tex]
Therefore, the maximum [tex]$y$[/tex]-value of \( g(x) \) is 2.
To summarize:
- \( g(x) \) achieves a maximum [tex]$y$[/tex]-value of 2 at \( x = 0 \).
- \( f(x) \) approaches a maximum [tex]$y$[/tex]-value of 2 as \( x \) approaches infinity but never actually reaches 2.
Now, let's match these conclusions with the given options:
A. \( f(x) \) and \( g(x) \) have equivalent maximum [tex]$y$[/tex]-values.
- This is incorrect because while \( g(x) \) reaches the value 2, \( f(x) \) only approaches 2.
B. \( g(x) \) has the largest possible [tex]$y$[/tex]-value.
- This is correct. \( g(x) \) indeed reaches a [tex]$y$[/tex]-value of 2, which is the largest possible value it attains.
C. \( f(x) \) has the largest possible [tex]$y$[/tex]-value.
- This is incorrect because \( f(x) \) does not reach the value 2, it only approaches it.
D. The maximum [tex]$y$[/tex]-value of \( f(x) \) approaches 2.
- This is correct. As \( x \) increases, \( f(x) \) approaches 2.
Thus, options B and D are correct, aligning with our detailed analysis.
1. Function \( f(x) = -5^x + 2 \):
- This is an exponential function where the base of the exponent is 5 and it is negative.
- As \( x \) increases, \( 5^x \) grows exponentially. Since this term is negative, \( -5^x \) also grows negatively.
- Therefore, as \( x \) approaches positive infinity, \( -5^x \) approaches negative infinity.
- Thus, \( f(x) \) will get closer and closer to \( 2 \) from below (since we are adding 2 to a very large negative number).
[tex]\[ \lim_{{x \to \infty}} f(x) = 2 \][/tex]
However, \( f(x) \) never actually reaches 2; it only approaches it.
2. Function \( g(x) = -5x^2 + 2 \):
- This is a quadratic function opening downwards because the coefficient of \( x^2 \) is negative.
- The vertex of this parabola represents the maximum point since the parabola opens downwards.
- The vertex of a parabola \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \).
- In this case, \( a = -5 \), \( b = 0 \), and \( c = 2 \), so the vertex is at \( x = 0 \).
[tex]\[ g(0) = -5(0)^2 + 2 = 2 \][/tex]
Therefore, the maximum [tex]$y$[/tex]-value of \( g(x) \) is 2.
To summarize:
- \( g(x) \) achieves a maximum [tex]$y$[/tex]-value of 2 at \( x = 0 \).
- \( f(x) \) approaches a maximum [tex]$y$[/tex]-value of 2 as \( x \) approaches infinity but never actually reaches 2.
Now, let's match these conclusions with the given options:
A. \( f(x) \) and \( g(x) \) have equivalent maximum [tex]$y$[/tex]-values.
- This is incorrect because while \( g(x) \) reaches the value 2, \( f(x) \) only approaches 2.
B. \( g(x) \) has the largest possible [tex]$y$[/tex]-value.
- This is correct. \( g(x) \) indeed reaches a [tex]$y$[/tex]-value of 2, which is the largest possible value it attains.
C. \( f(x) \) has the largest possible [tex]$y$[/tex]-value.
- This is incorrect because \( f(x) \) does not reach the value 2, it only approaches it.
D. The maximum [tex]$y$[/tex]-value of \( f(x) \) approaches 2.
- This is correct. As \( x \) increases, \( f(x) \) approaches 2.
Thus, options B and D are correct, aligning with our detailed analysis.
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