Algebra I & II · 30 min

Quadratics and the parabola

A quadratic is one curve, the parabola, wearing three different equations. Vertex form, factored form, standard form. The quadratic formula is not magic; it is completing the square, done once.

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Polynomials and degree

A polynomial is a sum of terms, each a number times a whole-number power of the variable:

anxn++a2x2+a1x+a0a_n x^n + \dots + a_2 x^2 + a_1 x + a_0

The degree is the largest power with a nonzero coefficient. Degree 11 is a line, y=a1x+a0y = a_1 x + a_0, last lesson’s object. This lesson is degree 22:

y=ax2+bx+cy = ax^2 + bx + c

That x2x^2 term bends the graph. A degree-2 polynomial is a quadratic, and its graph is a parabola: a single smooth U (or, when aa is negative, an upside-down U).

One curve, three equations

The parabola is the real object. The equation is just how you write it down, and there are three useful ways, each anchored to something different.

Standard form, y=ax2+bx+cy = ax^2 + bx + c: the default, good for reading off cc (the yy-intercept).

Vertex form, y=a(xh)2+ky = a(x - h)^2 + k: anchored at the vertex (h,k)(h, k), the parabola’s turning point. This form says “take the simplest parabola y=ax2y = ax^2 and shift it to sit at (h,k)(h, k).”

Factored form, y=a(xr1)(xr2)y = a(x - r_1)(x - r_2): anchored at the roots r1,r2r_1, r_2, the spots where the curve crosses the xx-axis.

-6-4-2246-4-2246
vertex y = 0.88(x)² - 2
standard y = 0.88x² - 2
factored y = 0.88(x + 1.51)(x - 1.51)
Δ = b² − 4ac = 7 two real roots

drag the yellow vertex to move the parabola; drag the blue handle (at x = h+2) to bend it. the three formulas are three views of one curve.

Drag the vertex to move the parabola; drag the steepness handle to bend it. All three equations update at once. They are never in conflict, because they are three descriptions of the one curve you’re holding.

The vertex sits at -b/(2a)

A parabola is perfectly symmetric about a vertical line through its vertex. That line sits at

x=b2ax = -\frac{b}{2a}

so the vertex’s xx-coordinate is b/(2a)-b/(2a), and its height is whatever yy you get by plugging that xx back in. Every parabola written in standard form hands you its turning point through this one formula.

Find the turning point

The quadratic y=x26x+5y = x^2 - 6x + 5 has a=1a = 1 and b=6b = -6.

Find the xx-coordinate of its vertex using x=b/(2a)x = -b/(2a). What is it?

Factoring: roots by sum and product

For a quadratic x2+bx+cx^2 + bx + c (with a=1a = 1), factoring means finding two numbers r1,r2r_1, r_2 so that

x2+bx+c=(xr1)(xr2)x^2 + bx + c = (x - r_1)(x - r_2)

Multiply the right side out and match: the two roots multiply to cc and add to b-b. That pairing is called Vieta’s relation, and it turns factoring into a small search: find two numbers with the right product and the right sum.

For x25x+6x^2 - 5x + 6: you need two numbers multiplying to 66 and adding to 55. That’s 22 and 33. So x25x+6=(x2)(x3)x^2 - 5x + 6 = (x - 2)(x - 3), and the roots are x=2x = 2 and x=3x = 3.

Zero product, not zero and

Once factored, finding the roots uses one property: if a product is zero, at least one factor is zero.

So (x2)(x3)=0(x - 2)(x - 3) = 0 forces x2=0x - 2 = 0 or x3=0x - 3 = 0. That’s an or, giving two separate roots, x=2x = 2 and x=3x = 3. It is not an and; no single xx makes both factors zero at once. The parabola crosses the axis at each root in turn, not at both simultaneously.

Factor and read the roots

Factor x25x+6x^2 - 5x + 6 into (xr1)(xr2)(x - r_1)(x - r_2) by finding two numbers with product 66 and sum 55.

What is the smaller of the two roots?

Completing the square, once

Factoring is lovely when the roots are tidy integers. They usually aren’t. So instead of guessing, derive a formula that always works, by completing the square.

Start from ax2+bx+c=0ax^2 + bx + c = 0. Divide by aa, move the constant, and add exactly the right number to make the left side a perfect square:

(x+b2a)2=b24ac4a2\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}

Take the square root of both sides and solve for xx. Nothing here is a trick; it’s the rewrites from lesson 1, applied with care. What falls out is the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

It is completing the square, frozen into a reusable result. Factoring is just this formula’s easy special case, the one where the roots happen to be nice.

The discriminant counts the roots

Look at the piece under the square root:

Δ=b24ac\Delta = b^2 - 4ac

This is the discriminant, and its sign alone tells you the root structure before you compute anything.

  • Δ>0\Delta > 0: the square root is a real nonzero number, the ±\pm splits it, two real roots. The parabola crosses the xx-axis twice.
  • Δ=0\Delta = 0: the square root is zero, the ±\pm does nothing, one repeated root. The parabola just kisses the axis at its vertex.
  • Δ<0\Delta < 0: the square root of a negative number is not real, no real roots. The parabola floats entirely above or below the axis.

You can see why in the widget: as the parabola slides off the xx-axis, the discriminant badge flips sign in step with the crossings disappearing.

Compute the discriminant

For x2+2x+5x^2 + 2x + 5, compute the discriminant Δ=b24ac\Delta = b^2 - 4ac.

What is Δ\Delta? (Its sign tells you this parabola never touches the xx-axis.)

Where this goes next

A quadratic is the simplest curved function, and the parabola is the shape of the simplest loss surface you can optimize. When module 10 teaches gradient descent, the first bowl it rolls a ball down is a parabola, and the minimum it hunts for is the vertex you’ve been dragging around.

g(f(x)) is one layer wired into the next. log(∏) = ∑(log) is why a million tiny probabilities don’t sink training. Everything that follows is those two facts at scale, and the parabola is the curve they first bend.

Lesson complete

Nice tinkering.