IDNLearn.com is your trusted platform for finding reliable answers. Join our community to receive prompt, thorough responses from knowledgeable experts.
Sagot :
Certainly! Let's convert both quadratic functions from their vertex form to standard form step by step and identify the [tex]\( y \)[/tex]-intercept.
### Problem 4: [tex]\( y = 5(x-2)^2 + 7 \)[/tex]
1. Start with the vertex form:
[tex]\[ y = 5(x-2)^2 + 7 \][/tex]
2. Expand the square:
[tex]\[ (x-2)^2 = x^2 - 4x + 4 \][/tex]
3. Distribute the coefficient [tex]\( a = 5 \)[/tex] through the expanded binomial:
[tex]\[ y = 5(x^2 - 4x + 4) + 7 \][/tex]
[tex]\[ y = 5x^2 - 20x + 20 + 7 \][/tex]
4. Combine like terms:
[tex]\[ y = 5x^2 - 20x + 27 \][/tex]
So, in standard form, the quadratic function is:
[tex]\[ y = 5x^2 - 20x + 27 \][/tex]
5. Identify the [tex]\( y \)[/tex]-intercept:
The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]. Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = 5(0)^2 - 20(0) + 27 = 27 \][/tex]
Therefore, the [tex]\( y \)[/tex]-intercept is [tex]\( (0, 27) \)[/tex].
### Problem 5: [tex]\( y = 3(x+1)^2 + 12 \)[/tex]
1. Start with the vertex form:
[tex]\[ y = 3(x+1)^2 + 12 \][/tex]
2. Expand the square:
[tex]\[ (x+1)^2 = x^2 + 2x + 1 \][/tex]
3. Distribute the coefficient [tex]\( a = 3 \)[/tex] through the expanded binomial:
[tex]\[ y = 3(x^2 + 2x + 1) + 12 \][/tex]
[tex]\[ y = 3x^2 + 6x + 3 + 12 \][/tex]
4. Combine like terms:
[tex]\[ y = 3x^2 + 6x + 15 \][/tex]
So, in standard form, the quadratic function is:
[tex]\[ y = 3x^2 + 6x + 15 \][/tex]
5. Identify the [tex]\( y \)[/tex]-intercept:
The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]. Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = 3(0)^2 + 6(0) + 15 = 15 \][/tex]
Therefore, the [tex]\( y \)[/tex]-intercept is [tex]\( (0, 15) \)[/tex].
### Summary of Functions and [tex]\( y \)[/tex]-Intercepts
1. [tex]\( y = 5x^2 - 20x + 27 \)[/tex] with [tex]\( y \)[/tex]-intercept [tex]\( (0, 27) \)[/tex].
2. [tex]\( y = 3x^2 + 6x + 15 \)[/tex] with [tex]\( y \)[/tex]-intercept [tex]\( (0, 15) \)[/tex].
### Graphing the Functions
To graph these functions:
1. Plot the [tex]\( y \)[/tex]-intercepts:
- For [tex]\( y = 5x^2 - 20x + 27 \)[/tex], plot the point [tex]\( (0, 27) \)[/tex].
- For [tex]\( y = 3x^2 + 6x + 15 \)[/tex], plot the point [tex]\( (0, 15) \)[/tex].
2. Plot the vertices and additional points to shape the parabolas:
- For [tex]\( y = 5(x - 2)^2 + 7 \)[/tex], the vertex is [tex]\( (2, 7) \)[/tex].
- For [tex]\( y = 3(x + 1)^2 + 12 \)[/tex], the vertex is [tex]\( (-1, 12) \)[/tex].
3. Draw smooth parabolas through these points:
- Both functions open upwards since the coefficients of [tex]\( x^2 \)[/tex] (5 and 3) are positive.
- The parabolas are symmetric about their respective vertices.
By following these steps, you can accurately graph the quadratic functions based on their standard forms and [tex]\( y \)[/tex]-intercepts.
### Problem 4: [tex]\( y = 5(x-2)^2 + 7 \)[/tex]
1. Start with the vertex form:
[tex]\[ y = 5(x-2)^2 + 7 \][/tex]
2. Expand the square:
[tex]\[ (x-2)^2 = x^2 - 4x + 4 \][/tex]
3. Distribute the coefficient [tex]\( a = 5 \)[/tex] through the expanded binomial:
[tex]\[ y = 5(x^2 - 4x + 4) + 7 \][/tex]
[tex]\[ y = 5x^2 - 20x + 20 + 7 \][/tex]
4. Combine like terms:
[tex]\[ y = 5x^2 - 20x + 27 \][/tex]
So, in standard form, the quadratic function is:
[tex]\[ y = 5x^2 - 20x + 27 \][/tex]
5. Identify the [tex]\( y \)[/tex]-intercept:
The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]. Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = 5(0)^2 - 20(0) + 27 = 27 \][/tex]
Therefore, the [tex]\( y \)[/tex]-intercept is [tex]\( (0, 27) \)[/tex].
### Problem 5: [tex]\( y = 3(x+1)^2 + 12 \)[/tex]
1. Start with the vertex form:
[tex]\[ y = 3(x+1)^2 + 12 \][/tex]
2. Expand the square:
[tex]\[ (x+1)^2 = x^2 + 2x + 1 \][/tex]
3. Distribute the coefficient [tex]\( a = 3 \)[/tex] through the expanded binomial:
[tex]\[ y = 3(x^2 + 2x + 1) + 12 \][/tex]
[tex]\[ y = 3x^2 + 6x + 3 + 12 \][/tex]
4. Combine like terms:
[tex]\[ y = 3x^2 + 6x + 15 \][/tex]
So, in standard form, the quadratic function is:
[tex]\[ y = 3x^2 + 6x + 15 \][/tex]
5. Identify the [tex]\( y \)[/tex]-intercept:
The [tex]\( y \)[/tex]-intercept occurs when [tex]\( x = 0 \)[/tex]. Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = 3(0)^2 + 6(0) + 15 = 15 \][/tex]
Therefore, the [tex]\( y \)[/tex]-intercept is [tex]\( (0, 15) \)[/tex].
### Summary of Functions and [tex]\( y \)[/tex]-Intercepts
1. [tex]\( y = 5x^2 - 20x + 27 \)[/tex] with [tex]\( y \)[/tex]-intercept [tex]\( (0, 27) \)[/tex].
2. [tex]\( y = 3x^2 + 6x + 15 \)[/tex] with [tex]\( y \)[/tex]-intercept [tex]\( (0, 15) \)[/tex].
### Graphing the Functions
To graph these functions:
1. Plot the [tex]\( y \)[/tex]-intercepts:
- For [tex]\( y = 5x^2 - 20x + 27 \)[/tex], plot the point [tex]\( (0, 27) \)[/tex].
- For [tex]\( y = 3x^2 + 6x + 15 \)[/tex], plot the point [tex]\( (0, 15) \)[/tex].
2. Plot the vertices and additional points to shape the parabolas:
- For [tex]\( y = 5(x - 2)^2 + 7 \)[/tex], the vertex is [tex]\( (2, 7) \)[/tex].
- For [tex]\( y = 3(x + 1)^2 + 12 \)[/tex], the vertex is [tex]\( (-1, 12) \)[/tex].
3. Draw smooth parabolas through these points:
- Both functions open upwards since the coefficients of [tex]\( x^2 \)[/tex] (5 and 3) are positive.
- The parabolas are symmetric about their respective vertices.
By following these steps, you can accurately graph the quadratic functions based on their standard forms and [tex]\( y \)[/tex]-intercepts.
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. IDNLearn.com is your reliable source for accurate answers. Thank you for visiting, and we hope to assist you again.
