Parabola worksheet

1. Vertex: (0, 2)

1. Find the middle line of the curve (the axis of symmetry). It passes through the vertex at y = −b ÷ 2a

   $\mathrm{yᵥ}$$=\frac{2}{2\times 0.5}$$=2$

2. Put that back into the equation (or use c − b² ÷ 4a) to get the vertex's x-coordinate

   $\mathrm{xᵥ}$$=2-\frac{{\left(\mathrm{-2}\right)}^2}{4\times 0.5}$$=0$

3. Find the focal length p = 1 ÷ 4a. It is the distance from the vertex to the focus, and from the vertex to the directrix

   $p$$=\frac{1}{4\times 0.5}$$=0.5$

4. Move a distance p from the vertex along the axis to reach the focus

   $\text{focus = (0.5, 2)}$

5. Move p the other way to reach the directrix, a line at a right angle to the axis

   $\text{x = −0.5}$

6. Write the vertex form, which shows the vertex directly

   $\text{x = 0.5(y − 2)²}$

7. So the vertex is

   $\text{(0, 2)}$

2. Vertex: (−1.5, 2.5)

1. Find the middle line of the curve (the axis of symmetry). It passes through the vertex at y = −b ÷ 2a

   $\mathrm{yᵥ}$$=\frac{2}{2\times 0.4}$$=2.5$

2. Put that back into the equation (or use c − b² ÷ 4a) to get the vertex's x-coordinate

   $\mathrm{xᵥ}$$=1-\frac{{\left(\mathrm{-2}\right)}^2}{4\times 0.4}$$=\mathrm{-1.5}$

3. Find the focal length p = 1 ÷ 4a. It is the distance from the vertex to the focus, and from the vertex to the directrix

   $p$$=\frac{1}{4\times 0.4}$$=0.625$

4. Move a distance p from the vertex along the axis to reach the focus

   $\text{focus = (−0.875, 2.5)}$

5. Move p the other way to reach the directrix, a line at a right angle to the axis

   $\text{x = −2.125}$

6. Write the vertex form, which shows the vertex directly

   $\text{x = 0.4(y − 2.5)² − 1.5}$

7. So the vertex is

   $\text{(−1.5, 2.5)}$

3. Vertex: (2, 0)

1. Find the middle line of the curve (the axis of symmetry). It passes through the vertex at x = −b ÷ 2a

   $\mathrm{xᵥ}$$=\frac{2}{2\times 0.5}$$=2$

2. Put that back into the equation (or use c − b² ÷ 4a) to get the vertex's y-coordinate

   $\mathrm{yᵥ}$$=2-\frac{{\left(\mathrm{-2}\right)}^2}{4\times 0.5}$$=0$

3. Find the focal length p = 1 ÷ 4a. It is the distance from the vertex to the focus, and from the vertex to the directrix

   $p$$=\frac{1}{4\times 0.5}$$=0.5$

4. Move a distance p from the vertex along the axis to reach the focus

   $\text{focus = (2, 0.5)}$

5. Move p the other way to reach the directrix, a line at a right angle to the axis

   $\text{y = −0.5}$

6. Write the vertex form, which shows the vertex directly

   $\text{y = 0.5(x − 2)²}$

7. So the vertex is

   $\text{(2, 0)}$

4. Vertex: (2.5, −1.5)

1. Find the middle line of the curve (the axis of symmetry). It passes through the vertex at x = −b ÷ 2a

   $\mathrm{xᵥ}$$=\frac{2}{2\times 0.4}$$=2.5$

2. Put that back into the equation (or use c − b² ÷ 4a) to get the vertex's y-coordinate

   $\mathrm{yᵥ}$$=1-\frac{{\left(\mathrm{-2}\right)}^2}{4\times 0.4}$$=\mathrm{-1.5}$

3. Find the focal length p = 1 ÷ 4a. It is the distance from the vertex to the focus, and from the vertex to the directrix

   $p$$=\frac{1}{4\times 0.4}$$=0.625$

4. Move a distance p from the vertex along the axis to reach the focus

   $\text{focus = (2.5, −0.875)}$

5. Move p the other way to reach the directrix, a line at a right angle to the axis

   $\text{y = −2.125}$

6. Write the vertex form, which shows the vertex directly

   $\text{y = 0.4(x − 2.5)² − 1.5}$

7. So the vertex is

   $\text{(2.5, −1.5)}$

5. Standard form: y = (1/6)x² − x + 4

1. Every point of a parabola is equally far from the focus and the directrix. So the vertex sits halfway between them

   $\mathrm{yᵥ}$$=\frac{4+1}{2}$$=2.5$

2. The focal length p is how far the vertex is from the focus, which is half the gap between focus and directrix. Its sign shows the direction

   $p$$=\frac{4-1}{2}$$=1.5$

3. Making the distance to the focus equal the distance to the directrix gives a = 1 ÷ 4p

   $a$$=\frac{1}{4\times 1.5}$$=0.166667$

4. Move a distance p from the vertex along the axis to reach the focus

   $\text{focus = (3, 4)}$

5. Move p the other way to reach the directrix, a line at a right angle to the axis

   $\text{y = 1}$

6. Write the vertex form, which shows the vertex directly

   $\text{y = (1/6)(x − 3)² + 2.5}$

7. Multiply out the vertex form to get the standard form

   $\text{y = (1/6)x² − x + 4}$

8. So the vertex is

   $\text{(3, 2.5)}$

6. Standard form: x = 0.5y² − 4y + 9.5

1. Every point of a parabola is equally far from the focus and the directrix. So the vertex sits halfway between them

   $\mathrm{xᵥ}$$=\frac{2+1}{2}$$=1.5$

2. The focal length p is how far the vertex is from the focus, which is half the gap between focus and directrix. Its sign shows the direction

   $p$$=\frac{2-1}{2}$$=0.5$

3. Making the distance to the focus equal the distance to the directrix gives a = 1 ÷ 4p

   $a$$=\frac{1}{4\times 0.5}$$=0.5$

4. Move a distance p from the vertex along the axis to reach the focus

   $\text{focus = (2, 4)}$

5. Move p the other way to reach the directrix, a line at a right angle to the axis

   $\text{x = 1}$

6. Write the vertex form, which shows the vertex directly

   $\text{x = 0.5(y − 4)² + 1.5}$

7. Multiply out the vertex form to get the standard form

   $\text{x = 0.5y² − 4y + 9.5}$

8. So the vertex is

   $\text{(1.5, 4)}$

7. Vertex: (−7/3, 10/3)

1. Find the middle line of the curve (the axis of symmetry). It passes through the vertex at y = −b ÷ 2a

   $\mathrm{yᵥ}$$=\frac{2}{2\times 0.3}$$=3.333333$

2. Put that back into the equation (or use c − b² ÷ 4a) to get the vertex's x-coordinate

   $\mathrm{xᵥ}$$=1-\frac{{\left(\mathrm{-2}\right)}^2}{4\times 0.3}$$=\mathrm{-2.333333}$

3. Find the focal length p = 1 ÷ 4a. It is the distance from the vertex to the focus, and from the vertex to the directrix

   $p$$=\frac{1}{4\times 0.3}$$=0.833333$

4. Move a distance p from the vertex along the axis to reach the focus

   $\text{focus = (−1.5, 10/3)}$

5. Move p the other way to reach the directrix, a line at a right angle to the axis

   $\text{x = −19/6}$

6. Write the vertex form, which shows the vertex directly

   $\text{x = 0.3(y − 10/3)² − 7/3}$

7. So the vertex is

   $\text{(−7/3, 10/3)}$

8. Vertex: (1.25, −0.25)

1. Find the middle line of the curve (the axis of symmetry). It passes through the vertex at x = −b ÷ 2a

   $\mathrm{xᵥ}$$=\frac{2}{2\times 0.8}$$=1.25$

2. Put that back into the equation (or use c − b² ÷ 4a) to get the vertex's y-coordinate

   $\mathrm{yᵥ}$$=1-\frac{{\left(\mathrm{-2}\right)}^2}{4\times 0.8}$$=\mathrm{-0.25}$

3. Find the focal length p = 1 ÷ 4a. It is the distance from the vertex to the focus, and from the vertex to the directrix

   $p$$=\frac{1}{4\times 0.8}$$=0.3125$

4. Move a distance p from the vertex along the axis to reach the focus

   $\text{focus = (1.25, 0.0625)}$

5. Move p the other way to reach the directrix, a line at a right angle to the axis

   $\text{y = −0.5625}$

6. Write the vertex form, which shows the vertex directly

   $\text{y = 0.8(x − 1.25)² − 0.25}$

7. So the vertex is

   $\text{(1.25, −0.25)}$

9. Standard form: y = (1/6)x² − (2/3)x + 19/6

1. Every point of a parabola is equally far from the focus and the directrix. So the vertex sits halfway between them

   $\mathrm{yᵥ}$$=\frac{4+1}{2}$$=2.5$

2. The focal length p is how far the vertex is from the focus, which is half the gap between focus and directrix. Its sign shows the direction

   $p$$=\frac{4-1}{2}$$=1.5$

3. Making the distance to the focus equal the distance to the directrix gives a = 1 ÷ 4p

   $a$$=\frac{1}{4\times 1.5}$$=0.166667$

4. Move a distance p from the vertex along the axis to reach the focus

   $\text{focus = (2, 4)}$

5. Move p the other way to reach the directrix, a line at a right angle to the axis

   $\text{y = 1}$

6. Write the vertex form, which shows the vertex directly

   $\text{y = (1/6)(x − 2)² + 2.5}$

7. Multiply out the vertex form to get the standard form

   $\text{y = (1/6)x² − (2/3)x + 19/6}$

8. So the vertex is

   $\text{(2, 2.5)}$

10. Vertex: (2, −1)

1. Find the middle line of the curve (the axis of symmetry). It passes through the vertex at x = −b ÷ 2a

   $\mathrm{xᵥ}$$=\frac{2}{2\times 0.5}$$=2$

2. Put that back into the equation (or use c − b² ÷ 4a) to get the vertex's y-coordinate

   $\mathrm{yᵥ}$$=1-\frac{{\left(\mathrm{-2}\right)}^2}{4\times 0.5}$$=\mathrm{-1}$

3. Find the focal length p = 1 ÷ 4a. It is the distance from the vertex to the focus, and from the vertex to the directrix

   $p$$=\frac{1}{4\times 0.5}$$=0.5$

4. Move a distance p from the vertex along the axis to reach the focus

   $\text{focus = (2, −0.5)}$

5. Move p the other way to reach the directrix, a line at a right angle to the axis

   $\text{y = −1.5}$

6. Write the vertex form, which shows the vertex directly

   $\text{y = 0.5(x − 2)² − 1}$

7. So the vertex is

   $\text{(2, −1)}$