To determine the points of intersection for the equations [tex]\( y = -2x + 3 \)[/tex] and [tex]\( y = x^2 - x + 1 \)[/tex], we must find the [tex]\( (x, y) \)[/tex] pairs that satisfy both equations simultaneously.
The equations are:
1. [tex]\( y = -2x + 3 \)[/tex]
2. [tex]\( y = x^2 - x + 1 \)[/tex]
Given our options, we need to evaluate each pair to see if they satisfy both equations:
1. Point [tex]\((0.5, 0.75)\)[/tex]:
- For [tex]\( y = -2x + 3 \)[/tex]:
[tex]\[ y = -2(0.5) + 3 = -1 + 3 = 2 \][/tex]
[tex]\[ y = 2 \neq 0.75 \][/tex]
This point does not satisfy the first equation.
- For [tex]\( y = x^2 - x + 1 \)[/tex]:
[tex]\[ y = (0.5)^2 - 0.5 + 1 = 0.25 - 0.5 + 1 = 0.75 \][/tex]
[tex]\[ y = 0.75 \][/tex]
This point satisfies the second equation but not the first.
2. Point [tex]\((1, 1)\)[/tex]:
- For [tex]\( y = -2x + 3 \)[/tex]:
[tex]\[ y = -2(1) + 3 = -2 + 3 = 1 \][/tex]
[tex]\[ y = 1 \][/tex]
This point satisfies the first equation.
- For [tex]\( y = x^2 - x + 1 \)[/tex]:
[tex]\[ y = (1)^2 - 1 + 1 = 1 - 1 + 1 = 1 \][/tex]
[tex]\[ y = 1 \][/tex]
This point satisfies the second equation.
3. Point [tex]\((-2, 7)\)[/tex]:
- For [tex]\( y = -2x + 3 \)[/tex]:
[tex]\[ y = -2(-2) + 3 = 4 + 3 = 7 \][/tex]
[tex]\[ y = 7 \][/tex]
This point satisfies the first equation.
- For [tex]\( y = x^2 - x + 1 \)[/tex]:
[tex]\[ y = (-2)^2 - (-2) + 1 = 4 + 2 + 1 = 7 \][/tex]
[tex]\[ y = 7 \][/tex]
This point satisfies the second equation.
4. Point [tex]\((0, 1)\)[/tex]:
- For [tex]\( y = -2x + 3 \)[/tex]:
[tex]\[ y = -2(0) + 3 = 3 \][/tex]
[tex]\[ y = 3 \neq 1 \][/tex]
This point does not satisfy the first equation.
- For [tex]\( y = x^2 - x + 1 \)[/tex]:
[tex]\[ y = (0)^2 - 0 + 1 = 1 \][/tex]
[tex]\[ y = 1 \][/tex]
This point satisfies the second equation but not the first.
Analyzing the points given in the problem:
- [tex]\((0.5, 0.75)\)[/tex] only satisfies the second equation.
- [tex]\((1, 1)\)[/tex] satisfies both equations.
- [tex]\((-2, 7)\)[/tex] satisfies both equations.
- [tex]\((0, 1)\)[/tex] only satisfies the second equation.
Therefore, the correct points of intersection for the equations [tex]\( y = -2x + 3 \)[/tex] and [tex]\( y = x^2 - x + 1 \)[/tex] are:
- [tex]\((-2, 7)\)[/tex]
- [tex]\((1, 1)\)[/tex]
The correct answer is:
[tex]\[ \boxed{(-2, 7) \text{ and } (1, 1)} \][/tex]
