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Graphing Quadratic Functions (Vertex & Axis of Symmetry) — Easy Worksheet 1
Grade 9 · MathSheets Name: ____________________ Date: __________
Write a vertex as an ordered pair like (2, -3), an axis of symmetry like x = 2, and x-intercepts like x = -1 or x = 5.
How to solve
For y = ax2 + bx + c, the axis of symmetry is x = -b2a. Substitute that x-value into the equation to get the y-coordinate of the vertex. If a > 0 the parabola opens up and the vertex is a minimum; if a < 0 it opens down and the vertex is a maximum.
Example
What is the vertex of the parabola?
Answer: (-1, -1)
Not quite. The answer is
💡 The parabola opens up, so its vertex is the lowest point. That turning point is at x = 0, y = −1, so the vertex is (0, −1).
✓ Correct!
Not quite. The answer is
💡 The parabola opens up, so its vertex is the lowest point. That turning point is at x = 0, y = −1, so the vertex is (0, −1).
💡 How to solve: Graphing Quadratic Functions (Vertex & Axis of Symmetry)
For y = ax2 + bx + c, the axis of symmetry is x = -b2a. Substitute that x-value into the equation to get the y-coordinate of the vertex. If a > 0 the parabola opens up and the vertex is a minimum; if a < 0 it opens down and the vertex is a maximum.
Try a similar one
What is the vertex of the parabola?
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 The axis of symmetry is x = −b2a. Here a = 2 and b = −16, so x = −−162 × 2 = 164 = 4. So the axis of symmetry is x = 4.
✓ Correct!
Not quite. The answer is
💡 The axis of symmetry is x = −b2a. Here a = 2 and b = −16, so x = −−162 × 2 = 164 = 4. So the axis of symmetry is x = 4.
💡 How to solve: Graphing Quadratic Functions (Vertex & Axis of Symmetry)
For y = ax2 + bx + c, the axis of symmetry is x = -b2a. Substitute that x-value into the equation to get the y-coordinate of the vertex. If a > 0 the parabola opens up and the vertex is a minimum; if a < 0 it opens down and the vertex is a maximum.
Try a similar one
What is the vertex of the parabola?
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 The axis of symmetry is x = −b2a. Here a = 2 and b = 8, so x = −82 × 2 = −84 = −2. Substitute x = −2 to get the y-value: y = 2(−2)2 + 8(−2) + 14 = 8 − 16 + 14 = 6. So the vertex is (−2, 6).
✓ Correct!
Not quite. The answer is
💡 The axis of symmetry is x = −b2a. Here a = 2 and b = 8, so x = −82 × 2 = −84 = −2. Substitute x = −2 to get the y-value: y = 2(−2)2 + 8(−2) + 14 = 8 − 16 + 14 = 6. So the vertex is (−2, 6).
💡 How to solve: Graphing Quadratic Functions (Vertex & Axis of Symmetry)
For y = ax2 + bx + c, the axis of symmetry is x = -b2a. Substitute that x-value into the equation to get the y-coordinate of the vertex. If a > 0 the parabola opens up and the vertex is a minimum; if a < 0 it opens down and the vertex is a maximum.
Try a similar one
What is the vertex of the parabola?
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 The x-intercepts are where the parabola crosses the x-axis (where y = 0). It crosses at x = 1 and x = 7, which are the same distance from the axis of symmetry x = 4. So x = 1 or x = 7.
✓ Correct!
Not quite. The answer is
💡 The x-intercepts are where the parabola crosses the x-axis (where y = 0). It crosses at x = 1 and x = 7, which are the same distance from the axis of symmetry x = 4. So x = 1 or x = 7.
💡 How to solve: Graphing Quadratic Functions (Vertex & Axis of Symmetry)
For y = ax2 + bx + c, the axis of symmetry is x = -b2a. Substitute that x-value into the equation to get the y-coordinate of the vertex. If a > 0 the parabola opens up and the vertex is a minimum; if a < 0 it opens down and the vertex is a maximum.
Try a similar one
What is the vertex of the parabola?
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 The parabola opens down, so its vertex is the highest point. That turning point is at x = 4, y = 2, so the vertex is (4, 2).
✓ Correct!
Not quite. The answer is
💡 The parabola opens down, so its vertex is the highest point. That turning point is at x = 4, y = 2, so the vertex is (4, 2).
💡 How to solve: Graphing Quadratic Functions (Vertex & Axis of Symmetry)
For y = ax2 + bx + c, the axis of symmetry is x = -b2a. Substitute that x-value into the equation to get the y-coordinate of the vertex. If a > 0 the parabola opens up and the vertex is a minimum; if a < 0 it opens down and the vertex is a maximum.
Try a similar one
What is the vertex of the parabola?
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 The axis of symmetry is x = −b2a. Here a = 1 and b = −2, so x = −−22 × 1 = 22 = 1. So the axis of symmetry is x = 1.
✓ Correct!
Not quite. The answer is
💡 The axis of symmetry is x = −b2a. Here a = 1 and b = −2, so x = −−22 × 1 = 22 = 1. So the axis of symmetry is x = 1.
💡 How to solve: Graphing Quadratic Functions (Vertex & Axis of Symmetry)
For y = ax2 + bx + c, the axis of symmetry is x = -b2a. Substitute that x-value into the equation to get the y-coordinate of the vertex. If a > 0 the parabola opens up and the vertex is a minimum; if a < 0 it opens down and the vertex is a maximum.
Try a similar one
What is the vertex of the parabola?
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 The axis of symmetry is x = −b2a. Here a = 2 and b = 16, so x = −162 × 2 = −164 = −4. Substitute x = −4 to get the y-value: y = 2(−4)2 + 16(−4) + 25 = 32 − 64 + 25 = −7. So the vertex is (−4, −7).
✓ Correct!
Not quite. The answer is
💡 The axis of symmetry is x = −b2a. Here a = 2 and b = 16, so x = −162 × 2 = −164 = −4. Substitute x = −4 to get the y-value: y = 2(−4)2 + 16(−4) + 25 = 32 − 64 + 25 = −7. So the vertex is (−4, −7).
💡 How to solve: Graphing Quadratic Functions (Vertex & Axis of Symmetry)
For y = ax2 + bx + c, the axis of symmetry is x = -b2a. Substitute that x-value into the equation to get the y-coordinate of the vertex. If a > 0 the parabola opens up and the vertex is a minimum; if a < 0 it opens down and the vertex is a maximum.
Try a similar one
What is the vertex of the parabola?
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
Not quite. The answer is
💡 The x-intercepts are where the parabola crosses the x-axis (where y = 0). It crosses at x = −1 and x = 3, which are the same distance from the axis of symmetry x = 1. So x = −1 or x = 3.
✓ Correct!
Not quite. The answer is
💡 The x-intercepts are where the parabola crosses the x-axis (where y = 0). It crosses at x = −1 and x = 3, which are the same distance from the axis of symmetry x = 1. So x = −1 or x = 3.
💡 How to solve: Graphing Quadratic Functions (Vertex & Axis of Symmetry)
For y = ax2 + bx + c, the axis of symmetry is x = -b2a. Substitute that x-value into the equation to get the y-coordinate of the vertex. If a > 0 the parabola opens up and the vertex is a minimum; if a < 0 it opens down and the vertex is a maximum.
Try a similar one
What is the vertex of the parabola?
✓ You've got it! Press Next to keep going. Not yet — the answer is . Read the steps above once more.
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Graphing Quadratic Functions (Vertex & Axis of Symmetry) — Easy Worksheet 1
Write a vertex as an ordered pair like (2, -3), an axis of symmetry like x = 2, and x-intercepts like x = -1 or x = 5.