Hodgson Schwartz
08/05/2023 · Junior High School
\( \begin{array}{ll}\text { 5. } f(x)=3 x^{2} & \text { 6. } f(x)=-x^{2}+1 \\ \text { 7. } f(x)=-x^{2}+4 x+1 & \text { 8. } f(x)=\frac{1}{2} x^{2}+6 x-7 \\ \text { 9. } y=(x+1)^{2} & \text { 10. } f(x)=(2 x-5)^{2}\end{array} \)
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It looks like you've provided a list of quadratic functions. If you need help with any specific tasks related to these functions, such as finding their vertex, graphing them, determining their roots, or analyzing their properties, please let me know! Here’s a brief overview of each function:
1. **Function 5:** \( f(x) = 3x^2 \)
- This is a parabola that opens upwards with vertex at the origin (0,0).
2. **Function 6:** \( f(x) = -x^2 + 1 \)
- This is a parabola that opens downwards with vertex at (0,1).
3. **Function 7:** \( f(x) = -x^2 + 4x + 1 \)
- This is a downward-opening parabola. To find the vertex, you can use the vertex formula \( x = -\frac{b}{2a} \).
4. **Function 8:** \( f(x) = \frac{1}{2}x^2 + 6x - 7 \)
- This is an upward-opening parabola. The vertex can also be found using the vertex formula.
5. **Function 9:** \( y = (x+1)^2 \)
- This is an upward-opening parabola with vertex at (-1,0).
6. **Function 10:** \( f(x) = (2x-5)^2 \)
- This is also an upward-opening parabola with vertex at \( \left(\frac{5}{2}, 0\right) \).
If you have a specific question or task in mind regarding these functions, please let me know!
Quick Answer
- Function 5: \( f(x) = 3x^2 \) opens upwards at (0,0).
- Function 6: \( f(x) = -x^2 + 1 \) opens downwards at (0,1).
- Function 7: \( f(x) = -x^2 + 4x + 1 \) opens downwards.
- Function 8: \( f(x) = \frac{1}{2}x^2 + 6x - 7 \) opens upwards.
- Function 9: \( y = (x+1)^2 \) opens upwards at (-1,0).
- Function 10: \( f(x) = (2x-5)^2 \) opens upwards at \( \left(\frac{5}{2}, 0\right) \).
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