Question
When graphing y = x2 – 14x + 24 and y = 20 – |x + 3|, which value would be the best minimum x- and minimum y-values for the viewing window so that all intercepts, minimums, maximums, and points of intersection are able to be seen as closely as possible?
–50
–45
–25
–20
–50
–45
–25
–20
Answers
Answered by
GPT-4o mini
To determine the best minimum x- and minimum y-values for the viewing window when graphing the equations \( y = x^2 - 14x + 24 \) and \( y = 20 - |x + 3| \), we need to analyze both functions.
1. **Finding the intercepts of \( y = x^2 - 14x + 24 \)**:
- For x-intercepts, set \( y = 0 \):
\[
x^2 - 14x + 24 = 0
\]
Factoring gives \( (x - 12)(x - 2) = 0 \), so the x-intercepts are \( x = 2 \) and \( x = 12 \).
- For the y-intercept, set \( x = 0 \):
\[
y = 0^2 - 14(0) + 24 = 24
\]
Thus, the y-intercept is \( y = 24 \).
2. **Finding the minimum value of \( y = x^2 - 14x + 24 \)**:
- The vertex of the parabola \( y = ax^2 + bx + c \) is located at \( x = -\frac{b}{2a} \):
\[
x = -\frac{-14}{2 \cdot 1} = 7
\]
- Substituting \( x = 7 \) back into the equation:
\[
y = 7^2 - 14(7) + 24 = 49 - 98 + 24 = -25
\]
So, the minimum value is \( y = -25 \) at \( x = 7 \).
3. **Finding the intercepts of \( y = 20 - |x + 3| \)**:
- For the x-intercepts, set \( y = 0 \):
\[
20 - |x + 3| = 0 \implies |x + 3| = 20
\]
This gives two equations:
\[
x + 3 = 20 \quad \Rightarrow \quad x = 17
\]
\[
x + 3 = -20 \quad \Rightarrow \quad x = -23
\]
Thus, the x-intercepts are \( x = 17 \) and \( x = -23 \).
- For the y-intercept, set \( x = 0 \):
\[
y = 20 - |0 + 3| = 20 - 3 = 17
\]
So, the y-intercept is \( y = 17 \).
4. **Finding the maximum value of \( y = 20 - |x + 3| \)**:
- The maximum value occurs at \( x = -3 \):
\[
y = 20 - | -3 + 3 | = 20 - 0 = 20
\]
5. **Summary of values**:
- x-intercepts of the quadratic: \( x = 2, 12 \)
- Minimum of the quadratic: \( y = -25 \) at \( x = 7 \)
- x-intercepts of \( y = 20 - |x + 3| \): \( x = -23, 17 \)
- Maximum of \( y = 20 - |x + 3| \): \( y = 20 \) at \( x = -3 \)
- y-intercepts: \( y = 24 \) for the quadratic, \( y = 17 \) for \( y = 20 - |x + 3| \)
6. **Determining the minimum viewing window**:
- For the x-values: The smallest x-intercept is \( -23 \) and the largest x-intercept is \( 17 \). Therefore, the minimum x-value can be \( -25 \) to ensure visibility.
- For the y-values: The minimum y-value is \( -25 \) from the quadratic, and the maximum y-value is \( 24 \). Therefore, the minimum y-value can also be \( -25 \).
Based on this analysis, the best minimum x-value is \( -25 \) which provides the required view to see the relevant points of interest in both graphs. Thus, the recommended values for the viewing window are:
- Minimum x-value: **–25**
- Minimum y-value: **–25**
Therefore, the answer is **–25** for the minimum x-value.
1. **Finding the intercepts of \( y = x^2 - 14x + 24 \)**:
- For x-intercepts, set \( y = 0 \):
\[
x^2 - 14x + 24 = 0
\]
Factoring gives \( (x - 12)(x - 2) = 0 \), so the x-intercepts are \( x = 2 \) and \( x = 12 \).
- For the y-intercept, set \( x = 0 \):
\[
y = 0^2 - 14(0) + 24 = 24
\]
Thus, the y-intercept is \( y = 24 \).
2. **Finding the minimum value of \( y = x^2 - 14x + 24 \)**:
- The vertex of the parabola \( y = ax^2 + bx + c \) is located at \( x = -\frac{b}{2a} \):
\[
x = -\frac{-14}{2 \cdot 1} = 7
\]
- Substituting \( x = 7 \) back into the equation:
\[
y = 7^2 - 14(7) + 24 = 49 - 98 + 24 = -25
\]
So, the minimum value is \( y = -25 \) at \( x = 7 \).
3. **Finding the intercepts of \( y = 20 - |x + 3| \)**:
- For the x-intercepts, set \( y = 0 \):
\[
20 - |x + 3| = 0 \implies |x + 3| = 20
\]
This gives two equations:
\[
x + 3 = 20 \quad \Rightarrow \quad x = 17
\]
\[
x + 3 = -20 \quad \Rightarrow \quad x = -23
\]
Thus, the x-intercepts are \( x = 17 \) and \( x = -23 \).
- For the y-intercept, set \( x = 0 \):
\[
y = 20 - |0 + 3| = 20 - 3 = 17
\]
So, the y-intercept is \( y = 17 \).
4. **Finding the maximum value of \( y = 20 - |x + 3| \)**:
- The maximum value occurs at \( x = -3 \):
\[
y = 20 - | -3 + 3 | = 20 - 0 = 20
\]
5. **Summary of values**:
- x-intercepts of the quadratic: \( x = 2, 12 \)
- Minimum of the quadratic: \( y = -25 \) at \( x = 7 \)
- x-intercepts of \( y = 20 - |x + 3| \): \( x = -23, 17 \)
- Maximum of \( y = 20 - |x + 3| \): \( y = 20 \) at \( x = -3 \)
- y-intercepts: \( y = 24 \) for the quadratic, \( y = 17 \) for \( y = 20 - |x + 3| \)
6. **Determining the minimum viewing window**:
- For the x-values: The smallest x-intercept is \( -23 \) and the largest x-intercept is \( 17 \). Therefore, the minimum x-value can be \( -25 \) to ensure visibility.
- For the y-values: The minimum y-value is \( -25 \) from the quadratic, and the maximum y-value is \( 24 \). Therefore, the minimum y-value can also be \( -25 \).
Based on this analysis, the best minimum x-value is \( -25 \) which provides the required view to see the relevant points of interest in both graphs. Thus, the recommended values for the viewing window are:
- Minimum x-value: **–25**
- Minimum y-value: **–25**
Therefore, the answer is **–25** for the minimum x-value.