Practice Maths

Graphing Parabolas

Key Ideas

Key Terms

Standard form
Y = ax² + bx + c. The sign of a determines direction; c is the y-intercept.
Vertex form
Y = a(x − h)² + k. The vertex is at (h, k).
Axis of symmetry
The vertical line x = −b ÷ (2a) — the parabola is a mirror image about this line.
Vertex x-coordinate
X = −b ÷ (2a). Substitute back into the equation to find the y-coordinate.
x-intercepts
Set y = 0 and solve ax² + bx + c = 0 (may have 0, 1, or 2 solutions).
y-intercept
Set x = 0; the y-intercept is always the constant c.
Feature How to find it Example: y = x² − 2x − 3
DirectionSign of a: positive ⇒ up; negative ⇒ downa = 1 > 0 ⇒ opens up
y-interceptSet x = 0; y = c(0, −3)
Axis of symmetryx = −b ÷ (2a)x = −(−2) ÷ 2 = x = 1
Vertexx from axis; y by substitutiony(1) = 1−2−3 = −4 ⇒ (1, −4)
x-interceptsSet y = 0; solve quadratic(x+1)(x−3)=0 ⇒ (−1, 0) and (3, 0)

Graph of y = x² − 2x − 3 with key features marked:

x y −2 −1 1 2 3 4 5 5 4 3 2 1 −1 −2 −3 −4 x = 1 (−1, 0) (3, 0) (0, −3) V(1, −4) y = x² − 2x − 3
x-intercepts y-intercept vertex — — axis of symmetry
Sketching Strategy — 5 steps: Determine direction (sign of a) → Mark y-intercept (0, c) → Find vertex using x = −b/(2a) → Find x-intercepts (if they exist) → Draw the axis of symmetry and sketch a smooth curve through all marked points.

Understanding the Role of a, b, and c

In y = ax² + bx + c, each coefficient plays a specific role:

  • a controls direction and width. If a > 0, the parabola opens upward (minimum turning point). If a < 0, it opens downward (maximum turning point). The larger |a| is, the narrower the parabola; the smaller |a|, the wider.
  • c is the y-intercept. Setting x = 0 gives y = c immediately. This is the simplest feature to read off.
  • b shifts the axis of symmetry. Combined with a, the axis is at x = −b/(2a). Note: b alone does not tell you how far the axis has shifted without also knowing a.

Why x = −b/(2a) Gives the Vertex

The quadratic formula gives x = (−b ± √Δ)/(2a). The two x-intercepts are symmetric about the axis. Their midpoint (the average) is [(−b + √Δ)/(2a) + (−b − √Δ)/(2a)] / 2 = [−2b/(2a)] / 2 = −b/(2a). So the axis of symmetry — and the vertex — always sits at x = −b/(2a), even when the x-intercepts are irrational or don’t exist.

Vertex Form: y = a(x − h)² + k

Vertex form makes the vertex (h, k) immediately obvious. To convert from standard form:

Method (completing the square): y = x² − 6x + 5  ⇒  y = (x² − 6x + 9) − 9 + 5  ⇒  y = (x − 3)² − 4. Vertex: (3, −4).

Shortcut: Find h = −b/(2a), then k = y(h) by substituting. This avoids completing the square for simple reading off the vertex.

To convert from vertex form to standard form: expand (x − h)² and collect terms. E.g. y = 2(x + 1)² − 3 = 2(x² + 2x + 1) − 3 = 2x² + 4x + 2 − 3 = 2x² + 4x − 1.

How Many x-Intercepts?

Use the discriminant Δ = b² − 4ac (from the previous lesson) to decide before solving:

  • Δ > 0: two x-intercepts. The parabola crosses the x-axis twice.
  • Δ = 0: one x-intercept (the vertex sits exactly on the x-axis).
  • Δ < 0: no x-intercepts. The parabola is entirely above (a > 0) or below (a < 0) the x-axis.

Finding the Equation from a Graph

If you know the x-intercepts at x = p and x = q, the equation has the form y = a(x − p)(x − q). Substitute a known point (such as the y-intercept) to find a.

If you know the vertex (h, k), use y = a(x − h)² + k and substitute one other point to find a.

Common Mistakes:
  • Reading off h incorrectly from vertex form: in y = (x − 3)², h = +3 (not −3). The formula is (x minus h).
  • Not substituting back to find the y-coordinate of the vertex after finding x = −b/(2a).
  • Forgetting that the y-intercept is always at x = 0, giving y = c in standard form.

Mastery Practice

  1. For each parabola, state the direction it opens, the y-intercept, the axis of symmetry, and the vertex. Fluency

      Equation Direction y-intercept Axis of symmetry Vertex
    (a)y = x² − 4x + 3    
    (b)y = −2x² + 8x − 5    
    (c)y = x² + 6x    
    (d)y = −x² + 4    

  2. Find the x-intercepts of each parabola, or state that there are none. Show your method. Fluency

    1. y = x² − 5x + 6
    2. y = x² − 4x + 4
    3. y = x² + 2x + 5
    4. y = 2x² − x − 3

  3. Complete a full feature analysis of y = x² − 4x + 3. Fluency

    1. Does the parabola open up or down? State the value of a.
    2. Write down the y-intercept.
    3. Find the equation of the axis of symmetry.
    4. Find the coordinates of the vertex.
    5. Find the x-intercepts (if any).

  4. Convert between standard form and vertex form. Fluency

    1. Write y = (x − 3)² − 4 in standard form.
    2. Write y = x² − 6x + 5 in vertex form. (Hint: complete the square.)
    3. Write y = 2(x + 1)² − 3 in standard form.
    4. Write y = x² − 4x + 1 in vertex form.

  5. Reading parabolas from equations. Understanding

    For each equation, describe the graph in words: state the direction, the vertex, whether it crosses the x-axis (and where), and the y-intercept.

    1. y = 2x²
    2. y = −(x − 2)² + 4
    3. y = x² − 1
    4. y = (x + 3)²

  6. Find the equation of each parabola. Understanding

    1. Vertex at (2, −3) and passes through (0, 1).
    2. x-intercepts at x = −1 and x = 5, and y-intercept = −5.
    3. Opens downward, vertex at (3, 7), and passes through (1, 3).

  7. Transformations of parabolas. Understanding

    Each part describes how one parabola relates to another. Describe the change in words (shift, reflection, or stretch/shrink).

    1. y = x²  →  y = (x − 3)²
    2. y = x²  →  y = x² + 5
    3. y = x²  →  y = −x²
    4. y = x²  →  y = 3x²
    5. y = x²  →  y = 2(x + 1)² − 3

  8. Ball trajectory. Understanding

    Physics. A ball is thrown upward. Its height h (metres) after t seconds is modelled by h = −t² + 4t + 5.
    1. What is the initial height of the ball (at t = 0)?
    2. Find the maximum height and the time at which it occurs.
    3. At what two times is the ball at a height of 8 m?
    4. When does the ball hit the ground? (Solve h = 0 and reject any invalid solution.)

  9. Finding an equation from three points. Problem Solving

    Algebraic derivation. A quadratic y = ax² + bx + c passes through the points (0, 6), (1, 4), and (3, 6).
    1. Substitute each point to write a system of three equations in a, b, and c.
    2. Use the substitution from (0, 6) to find c immediately.
    3. Solve the remaining two equations simultaneously to find a and b.
    4. Write the complete equation and find its vertex.

  10. Maximising area. Problem Solving

    Optimisation. A farmer wants to build a rectangular chicken run against an existing straight fence. He has 60 m of fencing for the three sides (the fence forms the fourth side).
    1. Let the width of the run (perpendicular to the fence) be x metres. Write an expression for the length in terms of x, then write the area A as a quadratic function of x.
    2. Show that A = −2x² + 60x. Find the axis of symmetry and the vertex of this parabola.
    3. State the dimensions that maximise the enclosed area and calculate the maximum area.
    4. What values of x are physically possible? Why must x be restricted to this range?
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