Skip to main content

Parabolas and the quadratic formula

Complete the square to write any quadratic in vertex form and read its lowest or highest point straight off, derive the quadratic formula, count the real zeros from the discriminant, and find the slope with the smallest squared error for the course’s data, all without calculus.
M01.3.3L0 Foundation25 minIn review

Before this lesson

M01.3.1 A function is a machine M01.3.2 Straight lines: slope and intercept

After this lesson you can

  • Complete the square to rewrite a quadratic in vertex form, showing every step.
  • Read the vertex form as the graph of x squared slid sideways and up or down, and say whether the vertex is the lowest or the highest point.
  • Derive the quadratic formula from the completed square, and count the real zeros from the sign of the discriminant.
  • Find the slope with the smallest squared error for the course’s data by locating the vertex of a parabola.

In review. The maths and code on this page are checked by automated tests on every change, but an independent reader has not signed it off yet. Spot a mistake or a confusing step? Tell us.

The question

Training a model means finding the settings with the smallest loss. Hold the course’s intercept at 0.5, and the mean squared error of the line \({\hat{y} = wx + 0.5}\) on the four data points is

\[ \begin{aligned} L(w) &= 7.5w^2 - 21w \\ &\quad + 14.75 . \end{aligned} \]

Which slope \({w}\) makes it smallest? Calculus can answer that, but for this curve plain algebra finds the exact answer.

From the last lesson

1. What is the slope of the line through \({(1, 2)}\) and \({(4, 6)}\)?

Show answer

Rise over run: \({\frac{6 - 2}{4 - 1} = \frac{4}{3}}\).

2. Where do \({y = 2x + 1}\) and \({y = -x + 7}\) cross?

Show answer

Set the outputs equal: \({2x + 1 = -x + 7}\), so \({3x = 6}\) and \({x = 2}\). Then \({y = 5}\): the point \({(2, 5)}\).

Intuition

The graph of \({x^2}\) is the U of lesson M01.3.1, with its lowest point at the origin. Every quadratic, a function \({ax^2 + bx + c}\), has a graph that is the same U, slid sideways, slid up or down, and perhaps stretched or turned upside down. That graph is called a parabola.

A coordinate plane with two parabolas. A thin sky-blue U labelled y equals x squared has its lowest point, a hollow dot, at the origin. A thick vermillion U labelled y equals x squared minus 4 x plus 3 has the same shape. A green arrow, labelled 2 right, 1 down, runs from the origin to its lowest point, ringed in yellow and labelled vertex (2, minus 1). A dashed vertical line passes through x equals 2, and two vermillion dots on the x-axis are labelled zeros 1 and 3. Beside the picture: the same curve, completed, y equals (x minus 2) squared minus 1; h equals 2: slide right 2; k equals minus 1: slide down 1; lowest point, the vertex: (h, k) equals (2, minus 1); outputs 0, the zeros: x equals 1 and x equals 3.
M01.3.3-F1

The parabola \({x^2 - 4x + 3}\) is the U of \({x^2}\) slid 2 to the right and 1 down. Written as \({(x - 2)^2 - 1}\), the slide can be read straight off the formula.

Take \({x^2 - 4x + 3}\). Written another way it is \({(x - 2)^2 - 1}\), and that form tells the whole story. The square \({(x - 2)^2}\) is never negative, and it is 0 only when \({x = 2}\). So the output is never below \({-1}\), and it equals \({-1}\) only at \({x = 2}\). The lowest point, the vertex, is \({(2, -1)}\). The outputs that equal 0, at \({x = 1}\) and \({x = 3}\), are the zeros: where the graph meets the \({x}\)-axis. They sit the same distance either side of the vertex.

This is how a loss is minimised without calculus: rewrite it so that a square shows, and read off where the square is 0. The rewriting has a name, completing the square.

Build it up

Step 1. Slide right. The graph of \({(x - h)^2}\) is the U moved \({h}\) to the right. Whatever \({x^2}\) gave at \({0}\), \({(x - h)^2}\) gives at \({h}\), so the vertex moves to \({(h, 0)}\).

Step 2. Slide up. Adding \({k}\) lifts every output by \({k}\): \({(x - h)^2 + k}\) has its vertex at \({(h, k)}\).

Step 3. Stretch or flip. Multiplying by \({a}\) scales every height. If \({a > 1}\) the U is narrower, if \({a}\) is between 0 and 1 it is wider, and if \({a < 0}\) it turns upside down, so that the vertex is the highest point. Together:

\[ y = a(x - h)^2 + k . \]

Step 4. Make a square. To reach that form from \({ax^2 + bx + c}\) we need one trick. The expression \({x^2 + 6x}\) is a square of side \({x}\) and two strips of \({3x}\), one on each side. Add the missing corner, \({3 \times 3 = 9}\), and it becomes a whole square of side \({x + 3}\).

Two panels. Left, titled x squared plus 6 x: a square and two strips: a blue square labelled x squared with sides x, an orange strip labelled 3 x to its right and another orange strip labelled 3 x below it, and a dashed empty square with a question mark in the corner between them, of side 3. Underneath: the 6 x is split into two strips of 3 x. Right, titled add the corner, 3 times 3 equals 9: the same pieces with the corner filled in yellow and labelled 9, making one large square with sides x plus 3. Underneath: x squared plus 6 x plus 9 equals (x plus 3) squared, a whole square.
M01.3.3-F2

Completing the square, as a picture. Split \({6x}\) into two strips of \({3x}\), and the square of half of 6 fills the gap.

In general, half of \({p}\), squared, completes \({x^2 + px}\). Expanding with the rule \({(u + v)^2} = {u^2 + 2uv + v^2}\) of lesson M01.1.2 checks it:

\[ \begin{aligned} \Big(x + \tfrac{p}{2}\Big)^2 &= x^2 + 2 \times \tfrac{p}{2}\,x + \tfrac{p^2}{4} \\ &= x^2 + px + \tfrac{p^2}{4} . \end{aligned} \]

The formal statement

NoteDefinition: quadratic, vertex, zeros

A quadratic function is \({f(x) = ax^2 + bx + c}\) with \({a \neq 0}\). Its graph is a parabola. A zero of \({f}\) is an input \({x}\) with \({f(x) = 0}\). The vertex is the lowest point of the graph if \({a > 0}\), and the highest point if \({a < 0}\).

Result 1: completing the square. Every step, for any \({a \neq 0}\). First take \({a}\) out of the two terms that contain \({x}\):

\[ f(x) = a\Big(x^2 + \tfrac{b}{a}x\Big) + c . \]

Half of \({\frac{b}{a}}\) is \({\frac{b}{2a}}\), and its square is \({\frac{b^2}{4a^2}}\). Add it and take it away inside the bracket, which changes nothing:

\[ \begin{aligned} f(x) = a\Big(&x^2 + \tfrac{b}{a}x + \tfrac{b^2}{4a^2} \\ &- \tfrac{b^2}{4a^2}\Big) + c . \end{aligned} \]

The first three terms in the bracket are a complete square, by Step 4 with \({p = \frac{b}{a}}\):

\[ f(x) = a\Big(\Big(x + \tfrac{b}{2a}\Big)^2 - \tfrac{b^2}{4a^2}\Big) + c . \]

Multiply the \({a}\) back in. It turns \({\frac{b^2}{4a^2}}\) into \({\frac{b^2}{4a}}\):

\[ f(x) = a\Big(x + \tfrac{b}{2a}\Big)^2 - \tfrac{b^2}{4a} + c . \]

So \({f(x) = a(x - h)^2 + k}\), with

\[ h = -\frac{b}{2a}, \qquad k = c - \frac{b^2}{4a} . \]

Result 2: the vertex. The square \({(x - h)^2}\) is never negative and is 0 only at \({x = h}\). If \({a > 0}\), then \({a(x - h)^2 \geq 0}\), so \({f(x) \geq k}\), with equality only at \({x = h}\): the lowest point is \({(h, k)}\). If \({a < 0}\), then \({a(x - h)^2 \leq 0}\) and \({(h, k)}\) is the highest point. Putting \({x = h}\) into \({f}\) gives \({f(h) = k}\), which is often the quickest way to find \({k}\).

Result 3: the quadratic formula. A zero solves \({a(x - h)^2 + k = 0}\). Take \({k}\) from both sides and divide by \({a}\):

\[ \begin{aligned} (x - h)^2 &= -\frac{k}{a} \\ &= \frac{b^2}{4a^2} - \frac{c}{a} \\ &= \frac{b^2 - 4ac}{4a^2} . \end{aligned} \]

The second line puts \({k = c - \frac{b^2}{4a}}\) in; the third writes both terms over \({4a^2}\). Call the top \({D = b^2 - 4ac}\), the discriminant. The bottom, \({4a^2}\), is positive.

  • If \({D > 0}\), \({x - h}\) is a square root of \({\frac{D}{4a^2}}\), so \({x - h = \pm\frac{\sqrt{D}}{2a}}\). (The square root of \({4a^2}\) is \({2a}\) or \({-2a}\), whichever is positive; with \({\pm}\) in front, both give the same pair of numbers.) Since \({h = -\frac{b}{2a}}\), the two zeros are

    \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} . \]

  • If \({D = 0}\), then \({(x - h)^2 = 0}\), so \({x = h}\): one zero, where the vertex touches the axis.

  • If \({D < 0}\), the right-hand side is negative, and no square is negative: there is no real zero. Lesson M01.9.4 returns to this case with new numbers.

Left, titled x squared minus 4 x plus c for c equals 3, 4, 5: three vermillion parabolas with their vertices one above another at x equals 2. The solid one, c equals 3, dips below the x-axis and meets it at the dots x equals 1 and x equals 3. The dashed one, c equals 4, touches the axis at x equals 2, ringed in yellow. The dotted one, c equals 5, stays above the axis. Right: a table with columns c, b squared minus 4 a c, and real zeros, with rows 3, 4, two: 1 and 3; 4, 0, one: 2; 5, minus 4, none. Below it: here a equals 1 and b equals minus 4, so b squared minus 4 a c equals 16 minus 4 c. Positive: two zeros. Zero: one. Negative: none.
M01.3.3-F3

The discriminant counts the real zeros. Raising \({c}\) lifts the U: it crosses the axis twice, then touches it, then misses it.

Go deeper: the sum and the product of the zeros

When \({D \geq 0}\) the zeros are \({h + \frac{\sqrt{D}}{2a}}\) and \({h - \frac{\sqrt{D}}{2a}}\). Their average is \({h}\): the zeros sit symmetrically about the vertex. Adding them gives \({2h = -\frac{b}{a}}\). Multiplying them, with \({(u + v)(u - v)} = {u^2 - v^2}\) and \({h^2 = \frac{b^2}{4a^2}}\), gives

\[ \begin{aligned} h^2 - \frac{D}{4a^2} &= \frac{b^2 - (b^2 - 4ac)}{4a^2} \\ &= \frac{4ac}{4a^2} = \frac{c}{a} . \end{aligned} \]

For \({x^2 - 4x + 3}\): \({1 + 3 = 4}\) and \({1 \times 3 = 3}\). A quick check on any answer.

Worked example

Part 1: \({x^2 - 4x + 3}\). Here \({a = 1}\), \({b = -4}\), \({c = 3}\). Half of \({-4}\) is \({-2}\), and \({(-2)^2 = 4}\). Add 4 and take 4 away:

\[ \begin{aligned} &x^2 - 4x + 3 \\ &\quad= (x^2 - 4x + 4) - 4 + 3 \\ &\quad= (x - 2)^2 - 1 . \end{aligned} \]

The vertex is \({(2, -1)}\), the lowest point because \({a = 1 > 0}\). The formulas agree: \({h = -\frac{-4}{2} = 2}\) and \({k = 3 - \frac{16}{4} = -1}\). The zeros solve \({(x - 2)^2 = 1}\), so \({x - 2 = 1}\) or \({x - 2 = -1}\): \({x = 3}\) or \({x = 1}\). The quadratic formula agrees too: \({D = 16 - 12 = 4}\), and \({x = \frac{4 \pm 2}{2}}\), which is 3 or 1.

Part 2: the course’s loss. With the intercept held at 0.5, the line predicts \({wx_i + 0.5}\), and the loss is the mean of the four squared misses. Four times the loss is their sum:

\[ \begin{aligned} 4L(w) &= (w - 1.5)^2 \\ &\quad+ (2w - 2.5)^2 \\ &\quad+ (3w - 4.5)^2 \\ &\quad+ (4w - 5.5)^2 . \end{aligned} \]

Each miss is \({wx_i + 0.5 - y_i}\); for the first point, \({w + 0.5 - 2 = w - 1.5}\). Expand each square with \({(u - v)^2} = {u^2 - 2uv + v^2}\) and add the columns:

square \({w^2}\) \({w}\) number
\({(w - 1.5)^2}\) 1 \({-3}\) 2.25
\({(2w - 2.5)^2}\) 4 \({-10}\) 6.25
\({(3w - 4.5)^2}\) 9 \({-27}\) 20.25
\({(4w - 5.5)^2}\) 16 \({-44}\) 30.25
sum 30 −84 59

So \({4L(w)} = {30w^2 - 84w + 59}\), and a quarter of that is \({L(w)} = {7.5w^2 - 21w + 14.75}\). Now the vertex:

\[ \begin{aligned} h &= \frac{21}{2 \times 7.5} = \frac{21}{15} = 1.4 , \\ k &= 14.75 - \frac{21^2}{4 \times 7.5} \\ &= 14.75 - 14.7 = 0.05 . \end{aligned} \]

So \({L(w)} = {7.5(w - 1.4)^2 + 0.05}\). The best slope is \({w = 1.4}\), with loss \({0.05}\): the course’s line \({\hat{y} = 1.4x + 0.5}\), found by algebra alone.

Left, titled Lines with b equals 0.5: the four blue data points and three vermillion lines starting from the same point on the y-axis, labelled w equals 1.8, w equals 1.4 (thick, close to all four points) and w equals 1. Right, titled Their loss: a parabola in w: a reddish-purple U-shaped curve of L against the slope w, labelled L of w equals 7.5 times (w minus 1.4) squared plus 0.05. Orange dots at w equals 1.0 and w equals 1.8 sit at the same height, joined by a dashed line labelled same loss, 1.25. A black star marks the lowest point, labelled (1.4, 0.05).
M01.3.3-F4

Each slope is a line on the left and a point on the right. The loss is a parabola in \({w}\), and its vertex, at \({w = 1.4}\), is the best slope.

The parabola has no zero, because its discriminant is negative:

\[ \begin{aligned} D &= 21^2 - 4 \times 7.5 \times 14.75 \\ &= 441 - 442.5 = -1.5 . \end{aligned} \]

That is no mistake. A loss of 0 would mean a line through all four points, and none exists.

Code

The vertex and the zeros, straight from Results 1 and 3. math.sqrt takes the square root; the if lines handle the three cases of the discriminant, and [] is an empty list.

import math

def vertex(a, b, c):
    h = -b / (2 * a)
    k = c - b * b / (4 * a)
    return h, k

def zeros(a, b, c):
    # the discriminant
    d = b * b - 4 * a * c
    if d < 0:
        # no real zero
        return []
    if d == 0:
        # one zero: the vertex
        return [-b / (2 * a)]
    root = math.sqrt(d)
    x1 = (-b - root) / (2 * a)
    x2 = (-b + root) / (2 * a)
    return [x1, x2]

print(vertex(1, -4, 3))
print(zeros(1, -4, 3))
h, k = vertex(7.5, -21, 14.75)
print(round(h, 4), round(k, 4))
print(zeros(7.5, -21, 14.75))
(2.0, -1.0)
[1.0, 3.0]
1.4 0.05
[]

The rounding only tidies the display: without it, k shows a tiny error in its last digits, because 14.7 has no exact binary form. Next, the loss computed straight from the data, beside the formula from the worked example. zip pairs each input with its target.

x = [1, 2, 3, 4]
y = [2, 3, 5, 6]

def loss(w, b=0.5):
    total = 0
    for xi, yi in zip(x, y):
        miss = w * xi + b - yi
        total += miss ** 2
    # the mean squared miss
    return total / len(x)

def formula(w):
    return 7.5 * w * w - 21 * w + 14.75

for w in [1.0, 1.4, 1.8]:
    direct = round(loss(w), 4)
    by_formula = round(formula(w), 4)
    print(w, direct, by_formula)
1.0 1.25 1.25
1.4 0.05 0.05
1.8 1.25 1.25

NumPy finds the zeros of a polynomial from its coefficients, highest power first.

import numpy as np

print(np.roots([1, -4, 3]))
print(np.roots([7.5, -21, 14.75]))
[3. 1.]
[1.4+0.08164966j 1.4-0.08164966j]

The first answer is the pair of zeros, 3 and 1. The second is printed with the letter j: those are not real numbers but the complex numbers of module M01.9, NumPy’s way of reporting that the loss has no real zero.

Explore

The function machine of this module is in lesson M01.3.1. Choose the rule \({x^2}\) and press Play: the U builds up point by point, and the outputs for \({-2}\) and \({2}\) land at the same height, as they must for a curve that is symmetric about its vertex.

Pitfalls

Two panels. Left, marked with a cross and titled Reading x minus 2 as a move left: a dashed reddish-purple parabola with its lowest point, a hollow dot, at x equals minus 2, labelled vertex at x equals minus 2?, and the note: but (x minus 2) squared minus 1 at x equals minus 2 gives (minus 4) squared minus 1 equals 15, not minus 1. Right, marked with a tick and titled The bracket is 0 at x equals 2: the vermillion parabola (x minus 2) squared minus 1 with its lowest point ringed in yellow and labelled vertex (2, minus 1), and the note: the bracket x minus 2 is 0 at x equals 2, so the square is smallest.
M01.3.3-F5

Wrong beside right. The minus sign in \({(x - 2)^2}\) slides the U to the right.

“\({(x - 2)^2 - 1}\) has its vertex at \({x = -2}\).”

The vertex is where the bracket is 0, and \({x - 2 = 0}\) at \({x = 2}\). A minus inside the bracket slides the graph right; a plus slides it left. Test the claim by putting \({x = -2}\) in: the output is \({(-4)^2 - 1 = 15}\), nowhere near the lowest value \({-1}\).

“Complete the square on \({2x^2 - 8x + 6}\) by halving \({-8}\).”

The trick works on \({x^2 + px}\), so take the 2 out first: \({2(x^2 - 4x) + 6}\). Now halve \({-4}\):

\[ \begin{aligned} &2\big((x - 2)^2 - 4\big) + 6 \\ &\quad= 2(x - 2)^2 - 8 + 6 \\ &\quad= 2(x - 2)^2 - 2 . \end{aligned} \]

The vertex is \({(2, -2)}\), as \({h = \frac{8}{4} = 2}\) and \({k = 6 - \frac{64}{8} = -2}\) confirm.

“Only \({\sqrt{b^2 - 4ac}}\) is divided by \({2a}\).”

The whole top, \({-b \pm \sqrt{b^2 - 4ac}}\), is divided by \({2a}\). For \({x^2 - 4x + 3}\) the right reading gives \({\frac{4 \pm 2}{2}}\), that is 3 and 1. Dividing only the root gives \({4 \pm 1}\), that is 5 and 3, and 5 is not a zero: \({25 - 20 + 3 = 8}\).

Check yourself

Recall

1. Give the vertex of \({y = -2(x - 1)^2 + 5}\). Is it the lowest or the highest point?

Show answer

\({(1, 5)}\). It is the highest point, because \({a = -2}\) is negative and the U is upside down.

2. What number completes the square for \({x^2 + 10x}\)?

Show answer

Half of 10, squared: \({25}\). Then \({x^2 + 10x + 25} = {(x + 5)^2}\).

3. State the quadratic formula for the zeros of \({ax^2 + bx + c}\).

Show answer

\({x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\).

4. How many real zeros has \({x^2 + 2x + 5}\)?

Show answer

None: \({b^2 - 4ac = 4 - 20} = {-16}\) is negative.

5. Why is \({(h, k)}\) the lowest point of \({a(x - h)^2 + k}\) when \({a > 0}\)?

Show answer

\({(x - h)^2}\) is never negative and is 0 only at \({x = h}\), so the output is at least \({k}\) and equals \({k}\) only at \({x = h}\).

Exercises

1. ★ Complete the square for \({x^2 + 6x + 5}\). Give the vertex and the zeros.

Show solution

Half of 6 is 3, and \({3^2 = 9}\):

\[ \begin{aligned} &x^2 + 6x + 5 \\ &\quad= (x^2 + 6x + 9) - 9 + 5 \\ &\quad= (x + 3)^2 - 4 . \end{aligned} \]

The vertex is \({(-3, -4)}\). The zeros solve \({(x + 3)^2 = 4}\), so \({x + 3 = 2}\) or \({x + 3 = -2}\): \({x = -1}\) or \({x = -5}\). Check: \({1 - 6 + 5 = 0}\) and \({25 - 30 + 5 = 0}\).

2. ★★ For \({f(x) = -x^2 + 4x + 5}\), find the vertex and say what kind of point it is, then find the zeros with the quadratic formula.

Show solution

Here \({a = -1}\), \({b = 4}\), \({c = 5}\). \({h = -\frac{4}{2 \times (-1)} = 2}\) and \({k = f(2) = -4 + 8 + 5} = {9}\). Since \({a < 0}\), \({(2, 9)}\) is the highest point. The discriminant is \({16 - 4 \times (-1) \times 5} = {36}\), so

\[ x = \frac{-4 \pm 6}{-2} , \]

which gives \({\frac{2}{-2} = -1}\) and \({\frac{-10}{-2} = 5}\). Check: \({-1 - 4 + 5 = 0}\) and \({-25 + 20 + 5 = 0}\). The zeros sit 3 either side of \({h = 2}\).

3. ★★★ Now hold the slope at \({w = 1.4}\) and let the intercept \({b}\) vary. Show that the course’s loss is \({L(b)} = {(b - 0.5)^2 + 0.05}\), and give the best intercept.

Show solution

The misses \({1.4x_i + b - y_i}\) are \({b - 0.6}\), \({b - 0.2}\), \({b - 0.8}\) and \({b - 0.4}\). Each square is \({b^2}\), minus twice the number times \({b}\), plus the number squared. The four numbers and their squares add up to

\[ \begin{aligned} &0.6 + 0.2 + 0.8 + 0.4 = 2 , \\ &0.36 + 0.04 \\ &\quad + 0.64 + 0.16 = 1.2 . \end{aligned} \]

So the four squares add up to \({4b^2 - 4b + 1.2}\), and a quarter of that is \({b^2 - b + 0.3}\). Half of \({-1}\) is \({-0.5}\), and \({0.5^2 = 0.25}\), so

\[ \begin{aligned} L(b) &= b^2 - b + 0.25 \\ &\quad - 0.25 + 0.3 \\ &= (b - 0.5)^2 + 0.05 . \end{aligned} \]

The best intercept is \({b = 0.5}\), with loss 0.05. Both knobs of the course’s line are at the bottom of their parabolas.

Summary card

A quadratic is the U of \({x^2}\), slid and stretched; completing the square shows where.

The anchor figure again: the U of y equals x squared and the same U slid 2 right and 1 down to its vertex (2, minus 1), the parabola x squared minus 4 x plus 3, with zeros at 1 and 3.
M01.3.3-F1

Slide the U to its vertex.

  • Vertex form. \({ax^2 + bx + c} = {a(x - h)^2 + k}\), with \({h = -\frac{b}{2a}}\) and \({k = f(h)}\).
  • Vertex. The lowest point if \({a > 0}\), the highest if \({a < 0}\).
  • Zeros. \({x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\): two, one or none as \({b^2 - 4ac}\) is positive, zero or negative.
  • The course’s loss. \({L(w)} = {7.5(w - 1.4)^2 + 0.05}\): best slope 1.4, loss 0.05.

Sources and next steps

Sources. Each was opened and checked against this lesson on 12 November 2026.

  • Jay Abramson and others, Algebra and Trigonometry 2e, OpenStax. Section 2.5: completing the square by adding the square of half the \({x}\) coefficient once the leading coefficient is 1, the quadratic formula derived by completing the square, and the discriminant table (zero gives one double solution, negative gives two complex solutions, so no real one). Section 5.1: the vertex form \({a(x - h)^2 + k}\) with \({h = -\frac{b}{2a}}\) and \({k = f(h)}\), a minimum when \({a > 0}\) and a maximum when \({a < 0}\), the graph as a shift of \({y = x^2}\), and zeros as \({x}\)-intercepts.
  • Google, Machine Learning Crash Course: Linear regression: Loss: the mean squared error as the average of the squared losses over the examples.
  • NumPy developers, numpy.roots, version 2.5: the roots of a polynomial from its coefficients, highest power first.

Next. M01.3.4 Chaining and undoing puts machines in a row and runs them backwards.

This lesson unlocks

M01.3.4 Chaining and undoing: composition and inverse M01.9.4 Quadratics with a negative discriminant

Your progress is saved in this browser only. No account, no tracking.
Back to top