Answer :
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.
