Answer :
Certainly! Let's analyze the given functions [tex]\( f(x) = x^2 + 4 \)[/tex] and [tex]\( g(y) = y^2 + 4 \)[/tex].
### Function [tex]\( f(x) = x^2 + 4 \)[/tex]:
1. Shape of the Graph: This is a quadratic function of the form [tex]\( f(x) = ax^2 + bx + c \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 0 \)[/tex], and [tex]\( c = 4 \)[/tex]. The graph of a quadratic function is a parabola.
2. Direction: Since [tex]\( a = 1 \)[/tex] is positive, the parabola opens upwards.
3. Vertex: The vertex of a parabola in the form [tex]\( ax^2 + bx + c \)[/tex] can be found at [tex]\( x = -\frac{b}{2a} \)[/tex]. Here, [tex]\( b = 0 \)[/tex], so the vertex is at [tex]\( x = 0 \)[/tex]. Plugging [tex]\( x = 0 \)[/tex] into the equation, [tex]\( f(0) = 0^2 + 4 = 4 \)[/tex]. Thus, the vertex is at [tex]\( (0, 4) \)[/tex].
4. Graph Shift: The [tex]\( +4 \)[/tex] indicates a vertical shift of the standard parabola [tex]\( y = x^2 \)[/tex] upwards by 4 units.
### Function [tex]\( g(y) = y^2 + 4 \)[/tex]:
1. Shape of the Graph: This is again a quadratic function similar to [tex]\( f(x) \)[/tex], just using [tex]\( y \)[/tex] instead of [tex]\( x \)[/tex]. Here, it's in the form [tex]\( g(y) = ay^2 + by + c \)[/tex], with the same values [tex]\( a = 1 \)[/tex], [tex]\( b = 0 \)[/tex], and [tex]\( c = 4 \)[/tex].
2. Direction: Since [tex]\( a = 1 \)[/tex] is positive, the parabola opens upwards.
3. Vertex: Because this is the same form as [tex]\( f(x) \)[/tex], the vertex will be at [tex]\( y = 0 \)[/tex] and, similarly, will be [tex]\( g(0) = 0^2 + 4 = 4 \)[/tex]. Thus, the vertex is at [tex]\( (0, 4) \)[/tex].
4. Graph Shift: The [tex]\( +4 \)[/tex] similarly indicates a vertical shift upwards by 4 units.
### Comparison:
- Both functions [tex]\( f(x) \)[/tex] and [tex]\( g(y) \)[/tex] are essentially the same in terms of their shape, direction, and vertical shift.
- Both graphs are parabolas that open upwards.
- Both have their vertices at [tex]\( (0, 4) \)[/tex] on their respective axes (x-axis for [tex]\( f(x) \)[/tex] and y-axis for [tex]\( g(y) \)[/tex]).
- The main difference is the variable used. [tex]\( f(x) \)[/tex] uses the x-axis as the independent variable, while [tex]\( g(y) \)[/tex] uses the y-axis as the independent variable. However, graphically, these differences do not affect the overall shape or position of the graphs relative to the coordinate system.
In conclusion, the graphical difference between [tex]\( f(x) = x^2 + 4 \)[/tex] and [tex]\( g(y) = y^2 + 4 \)[/tex] is purely in their notation, but not in their appearance on a coordinate plane. Both functions represent identical parabolas shifted upwards by 4 units.
### Function [tex]\( f(x) = x^2 + 4 \)[/tex]:
1. Shape of the Graph: This is a quadratic function of the form [tex]\( f(x) = ax^2 + bx + c \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 0 \)[/tex], and [tex]\( c = 4 \)[/tex]. The graph of a quadratic function is a parabola.
2. Direction: Since [tex]\( a = 1 \)[/tex] is positive, the parabola opens upwards.
3. Vertex: The vertex of a parabola in the form [tex]\( ax^2 + bx + c \)[/tex] can be found at [tex]\( x = -\frac{b}{2a} \)[/tex]. Here, [tex]\( b = 0 \)[/tex], so the vertex is at [tex]\( x = 0 \)[/tex]. Plugging [tex]\( x = 0 \)[/tex] into the equation, [tex]\( f(0) = 0^2 + 4 = 4 \)[/tex]. Thus, the vertex is at [tex]\( (0, 4) \)[/tex].
4. Graph Shift: The [tex]\( +4 \)[/tex] indicates a vertical shift of the standard parabola [tex]\( y = x^2 \)[/tex] upwards by 4 units.
### Function [tex]\( g(y) = y^2 + 4 \)[/tex]:
1. Shape of the Graph: This is again a quadratic function similar to [tex]\( f(x) \)[/tex], just using [tex]\( y \)[/tex] instead of [tex]\( x \)[/tex]. Here, it's in the form [tex]\( g(y) = ay^2 + by + c \)[/tex], with the same values [tex]\( a = 1 \)[/tex], [tex]\( b = 0 \)[/tex], and [tex]\( c = 4 \)[/tex].
2. Direction: Since [tex]\( a = 1 \)[/tex] is positive, the parabola opens upwards.
3. Vertex: Because this is the same form as [tex]\( f(x) \)[/tex], the vertex will be at [tex]\( y = 0 \)[/tex] and, similarly, will be [tex]\( g(0) = 0^2 + 4 = 4 \)[/tex]. Thus, the vertex is at [tex]\( (0, 4) \)[/tex].
4. Graph Shift: The [tex]\( +4 \)[/tex] similarly indicates a vertical shift upwards by 4 units.
### Comparison:
- Both functions [tex]\( f(x) \)[/tex] and [tex]\( g(y) \)[/tex] are essentially the same in terms of their shape, direction, and vertical shift.
- Both graphs are parabolas that open upwards.
- Both have their vertices at [tex]\( (0, 4) \)[/tex] on their respective axes (x-axis for [tex]\( f(x) \)[/tex] and y-axis for [tex]\( g(y) \)[/tex]).
- The main difference is the variable used. [tex]\( f(x) \)[/tex] uses the x-axis as the independent variable, while [tex]\( g(y) \)[/tex] uses the y-axis as the independent variable. However, graphically, these differences do not affect the overall shape or position of the graphs relative to the coordinate system.
In conclusion, the graphical difference between [tex]\( f(x) = x^2 + 4 \)[/tex] and [tex]\( g(y) = y^2 + 4 \)[/tex] is purely in their notation, but not in their appearance on a coordinate plane. Both functions represent identical parabolas shifted upwards by 4 units.
