Answer :
Certainly! Let's plot five points on the parabola given by the equation [tex]\( y = (x+3)^2 + 1 \)[/tex]. We'll choose five values for [tex]\( x \)[/tex], calculate the corresponding [tex]\( y \)[/tex] values, and then plot those points.
1. Choosing the first value for [tex]\( x \)[/tex]:
- Let [tex]\( x = -5 \)[/tex]
- Substitute [tex]\( x = -5 \)[/tex] into the equation:
[tex]\[ y = (-5 + 3)^2 + 1 = (-2)^2 + 1 = 4 + 1 = 5 \][/tex]
- The first point is [tex]\( (-5, 5) \)[/tex].
2. Choosing the second value for [tex]\( x \)[/tex]:
- Let [tex]\( x = -3 \)[/tex]
- Substitute [tex]\( x = -3 \)[/tex] into the equation:
[tex]\[ y = (-3 + 3)^2 + 1 = 0^2 + 1 = 0 + 1 = 1 \][/tex]
- The second point is [tex]\( (-3, 1) \)[/tex].
3. Choosing the third value for [tex]\( x \)[/tex]:
- Let [tex]\( x = -1 \)[/tex]
- Substitute [tex]\( x = -1 \)[/tex] into the equation:
[tex]\[ y = (-1 + 3)^2 + 1 = 2^2 + 1 = 4 + 1 = 5 \][/tex]
- The third point is [tex]\( (-1, 5) \)[/tex].
4. Choosing the fourth value for [tex]\( x \)[/tex]:
- Let [tex]\( x = 0 \)[/tex]
- Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = (0 + 3)^2 + 1 = 3^2 + 1 = 9 + 1 = 10 \][/tex]
- The fourth point is [tex]\( (0, 10) \)[/tex].
5. Choosing the fifth value for [tex]\( x \)[/tex]:
- Let [tex]\( x = 2 \)[/tex]
- Substitute [tex]\( x = 2 \)[/tex] into the equation:
[tex]\[ y = (2 + 3)^2 + 1 = 5^2 + 1 = 25 + 1 = 26 \][/tex]
- The fifth point is [tex]\( (2, 26) \)[/tex].
So, the five points on the parabola [tex]\( y = (x+3)^2 + 1 \)[/tex] are:
[tex]\[ (-5, 5), (-3, 1), (-1, 5), (0, 10), (2, 26) \][/tex]
These points can be plotted on a graph to visualize the parabola.
1. Choosing the first value for [tex]\( x \)[/tex]:
- Let [tex]\( x = -5 \)[/tex]
- Substitute [tex]\( x = -5 \)[/tex] into the equation:
[tex]\[ y = (-5 + 3)^2 + 1 = (-2)^2 + 1 = 4 + 1 = 5 \][/tex]
- The first point is [tex]\( (-5, 5) \)[/tex].
2. Choosing the second value for [tex]\( x \)[/tex]:
- Let [tex]\( x = -3 \)[/tex]
- Substitute [tex]\( x = -3 \)[/tex] into the equation:
[tex]\[ y = (-3 + 3)^2 + 1 = 0^2 + 1 = 0 + 1 = 1 \][/tex]
- The second point is [tex]\( (-3, 1) \)[/tex].
3. Choosing the third value for [tex]\( x \)[/tex]:
- Let [tex]\( x = -1 \)[/tex]
- Substitute [tex]\( x = -1 \)[/tex] into the equation:
[tex]\[ y = (-1 + 3)^2 + 1 = 2^2 + 1 = 4 + 1 = 5 \][/tex]
- The third point is [tex]\( (-1, 5) \)[/tex].
4. Choosing the fourth value for [tex]\( x \)[/tex]:
- Let [tex]\( x = 0 \)[/tex]
- Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = (0 + 3)^2 + 1 = 3^2 + 1 = 9 + 1 = 10 \][/tex]
- The fourth point is [tex]\( (0, 10) \)[/tex].
5. Choosing the fifth value for [tex]\( x \)[/tex]:
- Let [tex]\( x = 2 \)[/tex]
- Substitute [tex]\( x = 2 \)[/tex] into the equation:
[tex]\[ y = (2 + 3)^2 + 1 = 5^2 + 1 = 25 + 1 = 26 \][/tex]
- The fifth point is [tex]\( (2, 26) \)[/tex].
So, the five points on the parabola [tex]\( y = (x+3)^2 + 1 \)[/tex] are:
[tex]\[ (-5, 5), (-3, 1), (-1, 5), (0, 10), (2, 26) \][/tex]
These points can be plotted on a graph to visualize the parabola.
