Answer :
To graph the system of inequalities
[tex]\[ \begin{array}{l} y > x^2 - 2x + 3 \\ y \leq x + 5 \end{array} \][/tex]
we'll follow a step-by-step approach. Let's graph the inequalities one by one and then find the region that satisfies both conditions.
### Step 1: Graphing [tex]\( y > x^2 - 2x + 3 \)[/tex]
1. Graph the parabola [tex]\( y = x^2 - 2x + 3 \)[/tex]:
- This is a quadratic equation in standard form [tex]\( y = ax^2 + bx + c \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = 3 \)[/tex].
- First, find the vertex of the parabola using the vertex formula [tex]\( x = -\frac{b}{2a} = -\frac{-2}{2 \cdot 1} = 1 \)[/tex].
- Plug [tex]\( x = 1 \)[/tex] back into the equation to find the y-coordinate of the vertex: [tex]\( y = 1^2 - 2(1) + 3 = 1 - 2 + 3 = 2 \)[/tex].
- Therefore, the vertex is [tex]\( (1, 2) \)[/tex].
- Find the y-intercept by setting [tex]\( x = 0 \)[/tex]: [tex]\( y = 0^2 - 2(0) + 3 = 3 \)[/tex].
- The parabola opens upwards since the coefficient of [tex]\( x^2 \)[/tex] (which is 1) is positive.
2. Sketch the parabola:
- Plot the vertex at [tex]\( (1, 2) \)[/tex].
- Plot additional points on both sides of the vertex to get the shape of the parabola, such as:
- [tex]\( x = 0 \)[/tex], [tex]\( y = 3 \)[/tex] (the y-intercept).
- [tex]\( x = 2 \)[/tex], [tex]\( y = (2)^2 - 2(2) + 3 = 4 - 4 + 3 = 3 \)[/tex].
- Draw a dashed curve through these points because the inequality is strict ([tex]\( > \)[/tex]), meaning this line itself is not included in the region.
3. Shade the region above the parabola:
- Since [tex]\( y > x^2 - 2x + 3 \)[/tex], shade the area above the dashed parabola.
### Step 2: Graphing [tex]\( y \leq x + 5 \)[/tex]
1. Graph the line [tex]\( y = x + 5 \)[/tex]:
- This is a linear equation in slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m = 1 \)[/tex] (slope) and [tex]\( b = 5 \)[/tex] (y-intercept).
- Plot the y-intercept [tex]\( (0, 5) \)[/tex].
- Use the slope to find another point. Since the slope [tex]\( m = 1 \)[/tex], go up 1 unit and right 1 unit from the y-intercept to get the next point: [tex]\( (1, 6) \)[/tex].
2. Sketch the line:
- Draw a solid line through the points [tex]\( (0, 5) \)[/tex] and [tex]\( (1, 6) \)[/tex] because the inequality is non-strict ([tex]\( \leq \)[/tex]).
3. Shade the region below the line:
- Since [tex]\( y \leq x + 5 \)[/tex], shade the area below this line (including the line itself).
### Step 3: Determine the solution region
- The solution to the system of inequalities is the overlapping shaded region from both inequalities.
- Identify and highlight the intersection of the two shaded regions.
### Final Graph
1. Draw the parabola [tex]\( y = x^2 - 2x + 3 \)[/tex] with a dashed curve for [tex]\( y > x^2 - 2x + 3 \)[/tex] and shade above it.
2. Draw the line [tex]\( y = x + 5 \)[/tex] as a solid line for [tex]\( y \leq x + 5 \)[/tex] and shade below it.
3. The solution region is where the shading overlaps.
This overlapping region represents all [tex]\( (x, y) \)[/tex] pairs that satisfy both inequalities [tex]\( y > x^2 - 2x + 3 \)[/tex] and [tex]\( y \leq x + 5 \)[/tex].
[tex]\[ \begin{array}{l} y > x^2 - 2x + 3 \\ y \leq x + 5 \end{array} \][/tex]
we'll follow a step-by-step approach. Let's graph the inequalities one by one and then find the region that satisfies both conditions.
### Step 1: Graphing [tex]\( y > x^2 - 2x + 3 \)[/tex]
1. Graph the parabola [tex]\( y = x^2 - 2x + 3 \)[/tex]:
- This is a quadratic equation in standard form [tex]\( y = ax^2 + bx + c \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = 3 \)[/tex].
- First, find the vertex of the parabola using the vertex formula [tex]\( x = -\frac{b}{2a} = -\frac{-2}{2 \cdot 1} = 1 \)[/tex].
- Plug [tex]\( x = 1 \)[/tex] back into the equation to find the y-coordinate of the vertex: [tex]\( y = 1^2 - 2(1) + 3 = 1 - 2 + 3 = 2 \)[/tex].
- Therefore, the vertex is [tex]\( (1, 2) \)[/tex].
- Find the y-intercept by setting [tex]\( x = 0 \)[/tex]: [tex]\( y = 0^2 - 2(0) + 3 = 3 \)[/tex].
- The parabola opens upwards since the coefficient of [tex]\( x^2 \)[/tex] (which is 1) is positive.
2. Sketch the parabola:
- Plot the vertex at [tex]\( (1, 2) \)[/tex].
- Plot additional points on both sides of the vertex to get the shape of the parabola, such as:
- [tex]\( x = 0 \)[/tex], [tex]\( y = 3 \)[/tex] (the y-intercept).
- [tex]\( x = 2 \)[/tex], [tex]\( y = (2)^2 - 2(2) + 3 = 4 - 4 + 3 = 3 \)[/tex].
- Draw a dashed curve through these points because the inequality is strict ([tex]\( > \)[/tex]), meaning this line itself is not included in the region.
3. Shade the region above the parabola:
- Since [tex]\( y > x^2 - 2x + 3 \)[/tex], shade the area above the dashed parabola.
### Step 2: Graphing [tex]\( y \leq x + 5 \)[/tex]
1. Graph the line [tex]\( y = x + 5 \)[/tex]:
- This is a linear equation in slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m = 1 \)[/tex] (slope) and [tex]\( b = 5 \)[/tex] (y-intercept).
- Plot the y-intercept [tex]\( (0, 5) \)[/tex].
- Use the slope to find another point. Since the slope [tex]\( m = 1 \)[/tex], go up 1 unit and right 1 unit from the y-intercept to get the next point: [tex]\( (1, 6) \)[/tex].
2. Sketch the line:
- Draw a solid line through the points [tex]\( (0, 5) \)[/tex] and [tex]\( (1, 6) \)[/tex] because the inequality is non-strict ([tex]\( \leq \)[/tex]).
3. Shade the region below the line:
- Since [tex]\( y \leq x + 5 \)[/tex], shade the area below this line (including the line itself).
### Step 3: Determine the solution region
- The solution to the system of inequalities is the overlapping shaded region from both inequalities.
- Identify and highlight the intersection of the two shaded regions.
### Final Graph
1. Draw the parabola [tex]\( y = x^2 - 2x + 3 \)[/tex] with a dashed curve for [tex]\( y > x^2 - 2x + 3 \)[/tex] and shade above it.
2. Draw the line [tex]\( y = x + 5 \)[/tex] as a solid line for [tex]\( y \leq x + 5 \)[/tex] and shade below it.
3. The solution region is where the shading overlaps.
This overlapping region represents all [tex]\( (x, y) \)[/tex] pairs that satisfy both inequalities [tex]\( y > x^2 - 2x + 3 \)[/tex] and [tex]\( y \leq x + 5 \)[/tex].
