IDNLearn.com provides a seamless experience for finding and sharing answers. Find the answers you need quickly and accurately with help from our knowledgeable and dedicated community members.
Sagot :
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.
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Your questions deserve precise answers. Thank you for visiting IDNLearn.com, and see you again soon for more helpful information.
