Week 10 · Intermediate Algebra

10. Quadratic Functions & Vertex Form

120 min

Before you start

  • Factor monic and non-monic quadratics fluently
  • Read and write function notation f(x) and evaluate at specific values
  • Plot points on the Cartesian plane and read coordinates
  • Recognize the parabola as the graph of a degree-2 polynomial

By the end you'll be able to

  • Convert any standard-form quadratic to vertex form by completing the square
  • Compute the vertex using the formula x = -b/(2a) and back-substitute for y
  • Predict opening direction from the sign of the leading coefficient
  • Predict width from the magnitude of the leading coefficient
  • Identify axis of symmetry, vertex, and y-intercept of any parabola
Week 10 video coming soon
Read the lesson body below in the meantime.

Quadratic Functions & Vertex Form

A quadratic function f(x) = ax² + bx + c graphs as a parabola. Its shape and position are completely determined by a, b, and c.

Three forms of a quadratic

Form Expression What you read off easily
Standard ax² + bx + c y-intercept = c
Factored a(x − r)(x − s) x-intercepts (roots) = r, s
Vertex a(x − h)² + k Vertex = (h, k)

Each form is useful for a different question. To find the vertex, convert to vertex form.

Reading the parabola’s shape from a

  • a > 0: opens upward; vertex is a minimum.
  • a < 0: opens downward; vertex is a maximum.
  • |a| > 1: narrower than y = x².
  • |a| < 1: wider than y = x².

Finding the vertex

Two methods.

Method 1 — vertex formula

x = −b/(2a). Plug back into the function to get the y-coordinate.

For f(x) = 2x² − 8x + 7: x = 8/4 = 2. f(2) = 8 − 16 + 7 = −1. Vertex: (2, −1).

Method 2 — completing the square

Convert standard form to vertex form by adding and subtracting (b/2)².

For f(x) = x² − 6x + 5:

  1. Take half of −6: −3. Square it: 9.
  2. Rewrite: f(x) = (x² − 6x + 9) − 9 + 5 = (x − 3)² − 4.
  3. Vertex: (3, −4).

The trick is adding 9 and subtracting 9 simultaneously — it doesn’t change the function’s value, only its representation.

ML connection

MSE loss for linear regression J(θ) = (1/2m) Σ (h(x) − y)², viewed as a function of one parameter at a time, is a parabola. The minimum of that parabola is the optimal value of that parameter. Gradient descent walks downhill on this parabola; for linear regression with convex MSE, it’s guaranteed to reach the global minimum.

Gaussian probability density has a completed-square form in its exponent:

f(x) = (1 / σ√(2π)) · exp(−(x − μ)² / (2σ²))

The (x − μ)² is the same vertex-form structure you just studied. To convert any quadratic in the exponent of e into Gaussian form, you complete the square — exactly this week’s skill.

Common mistakes

These are the traps learners hit most often on this topic. Knowing them in advance is half the fix.

  • Forgetting to subtract back what you added when completing the square

    Adding inside the parens to form a perfect square requires subtracting the same value outside (or accounting for the leading coefficient). Otherwise the function value changes.

  • Reading the wrong sign for $h$ in vertex form

    has vertex , not . The sign in the vertex coordinate is the opposite of what’s inside the parentheses.

  • Confusing $|a|$ with $a$ for opening direction

    The sign of controls direction: opens up, opens down. controls width: larger → narrower parabola.

Practice problems

Try each on paper first. Click Show solution only after you've made a real attempt.

  1. Problem 1
    Find the vertex of via completing the square.
    Show solution

    . Vertex: (3, -4).

  2. Problem 2
    Find the vertex of via the formula.
    Show solution

    and . Vertex: (2, -1).

  3. Problem 3
    Convert to vertex form.
    Show solution

    .

  4. Problem 4
    Does open up or down?
    Show solution

    , so the parabola opens downward and the vertex is a maximum.

  5. Problem 5
    Why is MSE loss in linear regression a parabola in each parameter?
    Show solution

    MSE expands as a quadratic in each with positive leading coefficient, guaranteeing a parabola opening up. That convexity is why gradient descent converges.

  6. Problem 6
    Find the y-intercept and axis of symmetry of .
    Show solution

    y-intercept: . Axis of symmetry: .

  7. Problem 7
    Convert to vertex form.
    Show solution

    . Vertex: .

Practice quiz

  1. Question 1
    Vertex form of a quadratic:
  2. Question 2
    Find the vertex of f(x) = (x - 3)² + 5.
  3. Question 3
    If a > 0, the parabola opens:
  4. Question 4
    Vertex of f(x) = x² - 6x + 5 (after completing the square)?
  5. Question 5
    Convert f(x) = x² + 4x + 1 to vertex form.
  6. Question 6
    Find the vertex’s x-coordinate of f(x) = 2x² - 8x + 7. (Just the number.)
  7. Question 7
    ML connection: MSE J(θ) is a parabola in:
  8. Question 8
    Why does completing the square help in probability theory?
  9. Question 9
    Vertex of f(x) = -2(x - 1)² + 3?
  10. Reflection 10
    Why is MSE preferred over MAE for many ML tasks despite being more outlier-sensitive?

Week 10 recap

You translated between standard and vertex form, found vertices via the formula and via completing the square, and connected the parabola’s shape (open up versus down, narrow versus wide) to its leading coefficient. Three trap families fell: the add-and-subtract bookkeeping trap (forgetting to compensate when adding ), the inside-sign trap (reading vertex coordinates with the wrong sign), and the magnitude-vs-sign confusion (using when only the sign of matters for direction). Each outcome maps forward: convex MSE in linear regression inherits this parabolic structure; the Gaussian density’s exponent is a completed square; the vertex formula generalizes to first-order optimality conditions in calculus.

Coming next: Week 11 — Solving Quadratic Equations

Next week solves quadratic equations using three methods: factoring, the square root property, and the quadratic formula. You will know which method to reach for given the structure of the equation, and you will meet complex-number roots when the discriminant is negative. The choice between methods is itself a skill — the formula always works but factoring is faster when the quadratic factors nicely.

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