Answer :
To solve the problem, we need to analyze the given functions [tex]\( f \)[/tex] and [tex]\( g \)[/tex] and compare their characteristics based on their equations and properties.
### For Function [tex]\( f(x) \)[/tex]:
The equation of [tex]\( f(x) \)[/tex] is given by:
[tex]\[ f(x) = x^2 - 4x + 3 \][/tex]
First, let's find the [tex]\( y \)[/tex]-intercept of [tex]\( f \)[/tex]. The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 - 4(0) + 3 = 3 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept of function [tex]\( f \)[/tex] is [tex]\( 3 \)[/tex].
Next, we determine the vertex of the parabola defined by [tex]\( f(x) \)[/tex]. For a quadratic equation in the form [tex]\( ax^2 + bx + c \)[/tex], the [tex]\( x \)[/tex]-coordinate of the vertex is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex] and [tex]\( b = -4 \)[/tex]:
[tex]\[ x = -\frac{-4}{2(1)} = 2 \][/tex]
Substituting [tex]\( x = 2 \)[/tex] back into the function to find the [tex]\( y \)[/tex]-coordinate:
[tex]\[ f(2) = 2^2 - 4(2) + 3 = 4 - 8 + 3 = -1 \][/tex]
Thus, the vertex of [tex]\( f \)[/tex] is [tex]\( (2, -1) \)[/tex], which represents the minimum point on the downward parabola.
### For Function [tex]\( g(x) \)[/tex]:
The function [tex]\( g \)[/tex] is described as having a vertex at [tex]\( (1, 3) \)[/tex] and opens downwards. Therefore, its general form can be expressed as:
[tex]\[ g(x) = -a(x - 1)^2 + 3 \][/tex]
where [tex]\( a \)[/tex] is a positive constant.
To find the [tex]\( y \)[/tex]-intercept of [tex]\( g \)[/tex], we evaluate [tex]\( g(0) \)[/tex]:
[tex]\[ g(0) = -a(0 - 1)^2 + 3 \][/tex]
[tex]\[ g(0) = -a(1) + 3 \][/tex]
[tex]\[ g(0) = -a + 3 \][/tex]
Since [tex]\( a \)[/tex] is positive, [tex]\( -a + 3 \)[/tex] will always be less than 3.
### Comparing the [tex]\( y \)[/tex]-intercepts:
The [tex]\( y \)[/tex]-intercept of function [tex]\( f \)[/tex] is [tex]\( 3 \)[/tex], and the [tex]\( y \)[/tex]-intercept of function [tex]\( g \)[/tex] computed as [tex]\(-a + 3 \)[/tex], where [tex]\( a \)[/tex] is a positive constant, making the [tex]\( y \)[/tex]-intercept of [tex]\( g \)[/tex] less than [tex]\( 3 \)[/tex].
### Conclusion:
Given the comparisons:
- The [tex]\( y \)[/tex]-intercept of [tex]\( f \)[/tex] is [tex]\( 3 \)[/tex].
- The [tex]\( y \)[/tex]-intercept of [tex]\( g \)[/tex] is [tex]\( -a + 3 \)[/tex], where [tex]\( a \)[/tex] is positive, thus it is less than [tex]\( 3 \)[/tex].
Therefore, the statement that the [tex]\( y \)[/tex]-intercept of function [tex]\( f \)[/tex] is greater than the [tex]\( y \)[/tex]-intercept of function [tex]\( g \)[/tex] is true.
So, the correct answer is:
[tex]\[ \boxed{B} \][/tex]
### For Function [tex]\( f(x) \)[/tex]:
The equation of [tex]\( f(x) \)[/tex] is given by:
[tex]\[ f(x) = x^2 - 4x + 3 \][/tex]
First, let's find the [tex]\( y \)[/tex]-intercept of [tex]\( f \)[/tex]. The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 - 4(0) + 3 = 3 \][/tex]
Thus, the [tex]\( y \)[/tex]-intercept of function [tex]\( f \)[/tex] is [tex]\( 3 \)[/tex].
Next, we determine the vertex of the parabola defined by [tex]\( f(x) \)[/tex]. For a quadratic equation in the form [tex]\( ax^2 + bx + c \)[/tex], the [tex]\( x \)[/tex]-coordinate of the vertex is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex] and [tex]\( b = -4 \)[/tex]:
[tex]\[ x = -\frac{-4}{2(1)} = 2 \][/tex]
Substituting [tex]\( x = 2 \)[/tex] back into the function to find the [tex]\( y \)[/tex]-coordinate:
[tex]\[ f(2) = 2^2 - 4(2) + 3 = 4 - 8 + 3 = -1 \][/tex]
Thus, the vertex of [tex]\( f \)[/tex] is [tex]\( (2, -1) \)[/tex], which represents the minimum point on the downward parabola.
### For Function [tex]\( g(x) \)[/tex]:
The function [tex]\( g \)[/tex] is described as having a vertex at [tex]\( (1, 3) \)[/tex] and opens downwards. Therefore, its general form can be expressed as:
[tex]\[ g(x) = -a(x - 1)^2 + 3 \][/tex]
where [tex]\( a \)[/tex] is a positive constant.
To find the [tex]\( y \)[/tex]-intercept of [tex]\( g \)[/tex], we evaluate [tex]\( g(0) \)[/tex]:
[tex]\[ g(0) = -a(0 - 1)^2 + 3 \][/tex]
[tex]\[ g(0) = -a(1) + 3 \][/tex]
[tex]\[ g(0) = -a + 3 \][/tex]
Since [tex]\( a \)[/tex] is positive, [tex]\( -a + 3 \)[/tex] will always be less than 3.
### Comparing the [tex]\( y \)[/tex]-intercepts:
The [tex]\( y \)[/tex]-intercept of function [tex]\( f \)[/tex] is [tex]\( 3 \)[/tex], and the [tex]\( y \)[/tex]-intercept of function [tex]\( g \)[/tex] computed as [tex]\(-a + 3 \)[/tex], where [tex]\( a \)[/tex] is a positive constant, making the [tex]\( y \)[/tex]-intercept of [tex]\( g \)[/tex] less than [tex]\( 3 \)[/tex].
### Conclusion:
Given the comparisons:
- The [tex]\( y \)[/tex]-intercept of [tex]\( f \)[/tex] is [tex]\( 3 \)[/tex].
- The [tex]\( y \)[/tex]-intercept of [tex]\( g \)[/tex] is [tex]\( -a + 3 \)[/tex], where [tex]\( a \)[/tex] is positive, thus it is less than [tex]\( 3 \)[/tex].
Therefore, the statement that the [tex]\( y \)[/tex]-intercept of function [tex]\( f \)[/tex] is greater than the [tex]\( y \)[/tex]-intercept of function [tex]\( g \)[/tex] is true.
So, the correct answer is:
[tex]\[ \boxed{B} \][/tex]
