Answer :
Let's start with the original function [tex]\( f(x) = x^2 \)[/tex], which is a basic parabola opening upwards with its vertex at the origin [tex]\((0,0)\)[/tex].
The given function is [tex]\( h(x) = -(x-2)^2 \)[/tex]. We will determine the transformations step-by-step:
1. Horizontal Shift:
- In [tex]\( h(x) = -(x-2)^2 \)[/tex], we have [tex]\((x-2)\)[/tex] inside the square term. This represents a horizontal shift.
- Specifically, the graph of [tex]\( f(x) = x^2 \)[/tex] is shifted 2 units to the right.
- Thus, the new function after this shift is [tex]\( g(x) = (x-2)^2 \)[/tex]. The vertex of [tex]\( g(x) \)[/tex] is now at [tex]\((2,0)\)[/tex].
2. Reflection Over the x-axis:
- The negative sign outside the square in [tex]\( h(x) = -(x-2)^2 \)[/tex] causes a reflection over the x-axis.
- Reflecting [tex]\( g(x) = (x-2)^2 \)[/tex] over the x-axis changes the function to [tex]\( h(x) = -(x-2)^2 \)[/tex].
Now, let's look at the transformed function [tex]\( h(x) = -(x-2)^2 \)[/tex]:
- The vertex of the parabola [tex]\( g(x) = (x-2)^2 \)[/tex] at [tex]\((2,0)\)[/tex] is reflected to [tex]\((2,0)\)[/tex] in [tex]\( h(x) = -(x-2)^2 \)[/tex] because reflection preserves the x-coordinate of the vertex, but changes the sign of the y-coordinate.
- This reflection turns the parabola to open downward, and the values of [tex]\( h(x) \)[/tex] will be non-positive.
Let's examine specific values to understand the transformation.
- For [tex]\( x = 0 \)[/tex]:
- Original function [tex]\( f(0) = 0^2 = 0 \)[/tex]
- Horizontal shift: [tex]\( g(0) = (0-2)^2 = 4 \)[/tex]
- Reflection: [tex]\( h(0) = -(0-2)^2 = -4 \)[/tex]
- For [tex]\( x = 1 \)[/tex]:
- Original function [tex]\( f(1) = 1^2 = 1 \)[/tex]
- Horizontal shift: [tex]\( g(1) = (1-2)^2 = 1 \)[/tex]
- Reflection: [tex]\( h(1) = -(1-2)^2 = -1 \)[/tex]
- For [tex]\( x = 2 \)[/tex]:
- Original function [tex]\( f(2) = 2^2 = 4 \)[/tex]
- Horizontal shift: [tex]\( g(2) = (2-2)^2 = 0 \)[/tex]
- Reflection: [tex]\( h(2) = -(2-2)^2 = 0 \)[/tex]
- For [tex]\( x = 3 \)[/tex]:
- Original function [tex]\( f(3) = 3^2 = 9 \)[/tex]
- Horizontal shift: [tex]\( g(3) = (3-2)^2 = 1 \)[/tex]
- Reflection: [tex]\( h(3) = -(3-2)^2 = -1 \)[/tex]
- For [tex]\( x = 4 \)[/tex]:
- Original function [tex]\( f(4) = 4^2 = 16 \)[/tex]
- Horizontal shift: [tex]\( g(4) = (4-2)^2 = 4 \)[/tex]
- Reflection: [tex]\( h(4) = -(4-2)^2 = -4 \)[/tex]
Summarizing the above values:
- [tex]\( h(0) = -4 \)[/tex]
- [tex]\( h(1) = -1 \)[/tex]
- [tex]\( h(2) = 0 \)[/tex]
- [tex]\( h(3) = -1 \)[/tex]
- [tex]\( h(4) = -4 \)[/tex]
Thus, using these transformations, the graph of [tex]\( h(x) = -(x-2)^2 \)[/tex] can be visualized clearly with a vertex at [tex]\((2, 0)\)[/tex], opening downwards. The specific points [tex]\((0, -4), (1, -1), (2, 0), (3, -1), (4, -4)\)[/tex] illustrate the transformations.
The given function is [tex]\( h(x) = -(x-2)^2 \)[/tex]. We will determine the transformations step-by-step:
1. Horizontal Shift:
- In [tex]\( h(x) = -(x-2)^2 \)[/tex], we have [tex]\((x-2)\)[/tex] inside the square term. This represents a horizontal shift.
- Specifically, the graph of [tex]\( f(x) = x^2 \)[/tex] is shifted 2 units to the right.
- Thus, the new function after this shift is [tex]\( g(x) = (x-2)^2 \)[/tex]. The vertex of [tex]\( g(x) \)[/tex] is now at [tex]\((2,0)\)[/tex].
2. Reflection Over the x-axis:
- The negative sign outside the square in [tex]\( h(x) = -(x-2)^2 \)[/tex] causes a reflection over the x-axis.
- Reflecting [tex]\( g(x) = (x-2)^2 \)[/tex] over the x-axis changes the function to [tex]\( h(x) = -(x-2)^2 \)[/tex].
Now, let's look at the transformed function [tex]\( h(x) = -(x-2)^2 \)[/tex]:
- The vertex of the parabola [tex]\( g(x) = (x-2)^2 \)[/tex] at [tex]\((2,0)\)[/tex] is reflected to [tex]\((2,0)\)[/tex] in [tex]\( h(x) = -(x-2)^2 \)[/tex] because reflection preserves the x-coordinate of the vertex, but changes the sign of the y-coordinate.
- This reflection turns the parabola to open downward, and the values of [tex]\( h(x) \)[/tex] will be non-positive.
Let's examine specific values to understand the transformation.
- For [tex]\( x = 0 \)[/tex]:
- Original function [tex]\( f(0) = 0^2 = 0 \)[/tex]
- Horizontal shift: [tex]\( g(0) = (0-2)^2 = 4 \)[/tex]
- Reflection: [tex]\( h(0) = -(0-2)^2 = -4 \)[/tex]
- For [tex]\( x = 1 \)[/tex]:
- Original function [tex]\( f(1) = 1^2 = 1 \)[/tex]
- Horizontal shift: [tex]\( g(1) = (1-2)^2 = 1 \)[/tex]
- Reflection: [tex]\( h(1) = -(1-2)^2 = -1 \)[/tex]
- For [tex]\( x = 2 \)[/tex]:
- Original function [tex]\( f(2) = 2^2 = 4 \)[/tex]
- Horizontal shift: [tex]\( g(2) = (2-2)^2 = 0 \)[/tex]
- Reflection: [tex]\( h(2) = -(2-2)^2 = 0 \)[/tex]
- For [tex]\( x = 3 \)[/tex]:
- Original function [tex]\( f(3) = 3^2 = 9 \)[/tex]
- Horizontal shift: [tex]\( g(3) = (3-2)^2 = 1 \)[/tex]
- Reflection: [tex]\( h(3) = -(3-2)^2 = -1 \)[/tex]
- For [tex]\( x = 4 \)[/tex]:
- Original function [tex]\( f(4) = 4^2 = 16 \)[/tex]
- Horizontal shift: [tex]\( g(4) = (4-2)^2 = 4 \)[/tex]
- Reflection: [tex]\( h(4) = -(4-2)^2 = -4 \)[/tex]
Summarizing the above values:
- [tex]\( h(0) = -4 \)[/tex]
- [tex]\( h(1) = -1 \)[/tex]
- [tex]\( h(2) = 0 \)[/tex]
- [tex]\( h(3) = -1 \)[/tex]
- [tex]\( h(4) = -4 \)[/tex]
Thus, using these transformations, the graph of [tex]\( h(x) = -(x-2)^2 \)[/tex] can be visualized clearly with a vertex at [tex]\((2, 0)\)[/tex], opening downwards. The specific points [tex]\((0, -4), (1, -1), (2, 0), (3, -1), (4, -4)\)[/tex] illustrate the transformations.
