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Mathematical Foundations for ML

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Mathematical Foundations for ML · 95 lessons

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Curves given as $x$ in terms of $y$

Lines with $x$ as the subject

Explanation

Most equations we graph have $y$ as the subject, such as $y = mx + c$ or $y = x^2$ . When $y$ is on its own, we choose values of $x$ , compute the matching values of $y$ and plot the points.

An equation can equally have $x$ as the subject, giving $x$ in terms of $y$ . Then we choose values of $y$ and compute $x$ . The graph is still the set of all points that satisfy the equation.

A line with $x$ as the subject could always be rearranged to have $y$ as the subject, unless it is vertical. Here we work with equations that have $x$ as the subject, because some curves cannot be written as a single equation with $y$ as the subject. The simplest such equations describe straight lines.

The line

x = my + c

Theorem

For numbers $m$ and $c$ , the graph of

x = my + c

is a straight line.

It crosses the $x$ -axis at $(c, 0)$ .
It rises from left to right if $m > 0$ and falls if $m < 0$ .
If $m = 0$ , it is the vertical line $x = c$ .

The diagram below shows why, in four steps.

0,0

x

y

1

2

Step 1: Choose $y$ , compute $x$

For the line $x = 2 y + 1$ , we choose values of $y$ and compute $x$ . Choosing $y = - 2$ , $1$ and $2$ gives $x = - 3$ , $3$ and $5$ , so the points $(- 3, - 2)$ , $(3, 1)$ and $(5, 2)$ lie on the line.

Step 2: Where it crosses the $x$ -axis

Every point on the $x$ -axis has $y = 0$ . Putting $y = 0$ gives $x = 1$ , so the line crosses at $(1, 0)$ .

For $x = m y + c$ , putting $y = 0$ leaves $x = c$ , so the crossing point is $(c, 0)$ .

Step 3: Which way it leans

Each step of $1$ up in $y$ moves $x$ across by $m$ , here $2$ to the right. Since $m > 0$ , higher points lie further right, so the line rises from left to right, just as $y = m x + c$ does when $m > 0$ .

When $m < 0$ , higher points lie further left, so the line falls.

Step 4: Change $m$ and $c$

m = - 3

m = - \frac{5}{2}

m = - 2

m = - \frac{3}{2}

m = - 1

m = - \frac{1}{2}

m = 0

m = \frac{1}{2}

m = 1

m = \frac{3}{2}

m = 2

m = \frac{5}{2}

m = 3

c = - 3

c = - 2

c = - 1

c = 0

c = 1

c = 2

c = 3

x = - 3 y - 3

x = - 3 y - 2

x = - 3 y - 1

x = - 3 y

x = - 3 y + 1

x = - 3 y + 2

x = - 3 y + 3

x = - \frac{5}{2} y - 3

x = - \frac{5}{2} y - 2

x = - \frac{5}{2} y - 1

x = - \frac{5}{2} y

x = - \frac{5}{2} y + 1

x = - \frac{5}{2} y + 2

x = - \frac{5}{2} y + 3

x = - 2 y - 3

x = - 2 y - 2

x = - 2 y - 1

x = - 2 y

x = - 2 y + 1

x = - 2 y + 2

x = - 2 y + 3

x = - \frac{3}{2} y - 3

x = - \frac{3}{2} y - 2

x = - \frac{3}{2} y - 1

x = - \frac{3}{2} y

x = - \frac{3}{2} y + 1

x = - \frac{3}{2} y + 2

x = - \frac{3}{2} y + 3

x = - y - 3

x = - y - 2

x = - y - 1

x = - y

x = - y + 1

x = - y + 2

x = - y + 3

x = - \frac{1}{2} y - 3

x = - \frac{1}{2} y - 2

x = - \frac{1}{2} y - 1

x = - \frac{1}{2} y

x = - \frac{1}{2} y + 1

x = - \frac{1}{2} y + 2

x = - \frac{1}{2} y + 3

x = - 3

x = - 2

x = - 1

x = 0

x = 1

x = 2

x = 3

x = \frac{1}{2} y - 3

x = \frac{1}{2} y - 2

x = \frac{1}{2} y - 1

x = \frac{1}{2} y

x = \frac{1}{2} y + 1

x = \frac{1}{2} y + 2

x = \frac{1}{2} y + 3

x = y - 3

x = y - 2

x = y - 1

x = y

x = y + 1

x = y + 2

x = y + 3

x = \frac{3}{2} y - 3

x = \frac{3}{2} y - 2

x = \frac{3}{2} y - 1

x = \frac{3}{2} y

x = \frac{3}{2} y + 1

x = \frac{3}{2} y + 2

x = \frac{3}{2} y + 3

x = 2 y - 3

x = 2 y - 2

x = 2 y - 1

x = 2 y

x = 2 y + 1

x = 2 y + 2

x = 2 y + 3

x = \frac{5}{2} y - 3

x = \frac{5}{2} y - 2

x = \frac{5}{2} y - 1

x = \frac{5}{2} y

x = \frac{5}{2} y + 1

x = \frac{5}{2} y + 2

x = \frac{5}{2} y + 3

x = 3 y - 3

x = 3 y - 2

x = 3 y - 1

x = 3 y

x = 3 y + 1

x = 3 y + 2

x = 3 y + 3

The further $m$ is from $0$ , the further the line runs across for each step up, so the flatter it is. For $y = m x + c$ it is the other way round.

Example

Graph the equation $x = 2y - 4$ . Find where it crosses the $x$ -axis, say which way it leans, and plot it from a table of values.

Solution

Putting $y = 0$ gives $x = -4$ , so the line crosses the $x$ -axis at $(-4, 0)$ .

Here $m = 2$ is positive, so the line rises from left to right.

We choose three values of $y$ and compute $x = 2y - 4$ for each.

$y$	$0$	$1$	$2$
$x$	$-4$	$-2$	$0$

We plot the points $(-4, 0)$ , $(-2, 1)$ and $(0, 2)$ and join them with a straight line.

0,0

x

y

(- 4, 0)

(- 2, 1)

(0, 2)

Practice questions

4 questions

Four graphs are shown below.

0,0

−1

−2

−3

−1

−2

−3

A

−1

−2

−3

−1

−2

−3

B

−1

−2

−3

−1

−2

−3

C

−1

−2

−3

−1

−2

−3

D

Which one is the graph of $x = 2y + 1$ ?

Select the correct answer:

Graph $A$

Graph $B$

Graph $C$

Graph $D$

+ 3 more questions

Sideways parabolas

Explanation

After lines, the next simplest equations with $x$ as the subject are those that square $y$ . The parabola $y = x^2$ opens upwards. Swapping the roles of $x$ and $y$ gives $x = y^2$ , the same parabola turned on its side so that it opens to the right.

We graph it in the same way as a line with $x$ as the subject: we choose values of $y$ and compute $x$ .

Sideways parabola

Definition

For numbers $a$ and $c$ with $a \neq 0$ , the graph of

x = ay^2 + c

is a sideways parabola.

Which way a sideways parabola opens

Theorem

The sideways parabola $x = ay^2 + c$ opens to the right if $a > 0$ and opens to the left if $a < 0$ .
Its turning point, the leftmost or rightmost point, is $(c, 0)$ .

0,0

x

y

The blue curve is the graph of $x = a y^{2} + c$ , and the red point is its turning point. Move the sliders to change $a$ and $c$ .

a = - 2

a = - \frac{3}{2}

a = - 1

a = - \frac{1}{2}

a = 0

a = \frac{1}{2}

a = 1

a = \frac{3}{2}

a = 2

c = - 3

c = - 2

c = - 1

c = 0

c = 1

c = 2

c = 3

x = - 2 y^{2} - 3

x = - 2 y^{2} - 2

x = - 2 y^{2} - 1

x = - 2 y^{2}

x = - 2 y^{2} + 1

x = - 2 y^{2} + 2

x = - 2 y^{2} + 3

x = - \frac{3}{2} y^{2} - 3

x = - \frac{3}{2} y^{2} - 2

x = - \frac{3}{2} y^{2} - 1

x = - \frac{3}{2} y^{2}

x = - \frac{3}{2} y^{2} + 1

x = - \frac{3}{2} y^{2} + 2

x = - \frac{3}{2} y^{2} + 3

x = - y^{2} - 3

x = - y^{2} - 2

x = - y^{2} - 1

x = - y^{2}

x = - y^{2} + 1

x = - y^{2} + 2

x = - y^{2} + 3

x = - \frac{1}{2} y^{2} - 3

x = - \frac{1}{2} y^{2} - 2

x = - \frac{1}{2} y^{2} - 1

x = - \frac{1}{2} y^{2}

x = - \frac{1}{2} y^{2} + 1

x = - \frac{1}{2} y^{2} + 2

x = - \frac{1}{2} y^{2} + 3

x = - 3

x = - 2

x = - 1

x = 0

x = 1

x = 2

x = 3

x = \frac{1}{2} y^{2} - 3

x = \frac{1}{2} y^{2} - 2

x = \frac{1}{2} y^{2} - 1

x = \frac{1}{2} y^{2}

x = \frac{1}{2} y^{2} + 1

x = \frac{1}{2} y^{2} + 2

x = \frac{1}{2} y^{2} + 3

x = y^{2} - 3

x = y^{2} - 2

x = y^{2} - 1

x = y^{2}

x = y^{2} + 1

x = y^{2} + 2

x = y^{2} + 3

x = \frac{3}{2} y^{2} - 3

x = \frac{3}{2} y^{2} - 2

x = \frac{3}{2} y^{2} - 1

x = \frac{3}{2} y^{2}

x = \frac{3}{2} y^{2} + 1

x = \frac{3}{2} y^{2} + 2

x = \frac{3}{2} y^{2} + 3

x = 2 y^{2} - 3

x = 2 y^{2} - 2

x = 2 y^{2} - 1

x = 2 y^{2}

x = 2 y^{2} + 1

x = 2 y^{2} + 2

x = 2 y^{2} + 3

It opens to the left and its turning point is

(- 3, 0)

It opens to the left and its turning point is

(- 2, 0)

It opens to the left and its turning point is

(- 1, 0)

It opens to the left and its turning point is

(0, 0)

It opens to the left and its turning point is

(1, 0)

It opens to the left and its turning point is

(2, 0)

It opens to the left and its turning point is

(3, 0)

It opens to the left and its turning point is

(- 3, 0)

It opens to the left and its turning point is

(- 2, 0)

It opens to the left and its turning point is

(- 1, 0)

It opens to the left and its turning point is

(0, 0)

It opens to the left and its turning point is

(1, 0)

It opens to the left and its turning point is

(2, 0)

It opens to the left and its turning point is

(3, 0)

It opens to the left and its turning point is

(- 3, 0)

It opens to the left and its turning point is

(- 2, 0)

It opens to the left and its turning point is

(- 1, 0)

It opens to the left and its turning point is

(0, 0)

It opens to the left and its turning point is

(1, 0)

It opens to the left and its turning point is

(2, 0)

It opens to the left and its turning point is

(3, 0)

It opens to the left and its turning point is

(- 3, 0)

It opens to the left and its turning point is

(- 2, 0)

It opens to the left and its turning point is

(- 1, 0)

It opens to the left and its turning point is

(0, 0)

It opens to the left and its turning point is

(1, 0)

It opens to the left and its turning point is

(2, 0)

It opens to the left and its turning point is

(3, 0)

This is the vertical line

x = - 3

, not a parabola, so a sideways parabola needs

a \neq 0

This is the vertical line

x = - 2

, not a parabola, so a sideways parabola needs

a \neq 0

This is the vertical line

x = - 1

, not a parabola, so a sideways parabola needs

a \neq 0

This is the vertical line

x = 0

, not a parabola, so a sideways parabola needs

a \neq 0

This is the vertical line

x = 1

, not a parabola, so a sideways parabola needs

a \neq 0

This is the vertical line

x = 2

, not a parabola, so a sideways parabola needs

a \neq 0

This is the vertical line

x = 3

, not a parabola, so a sideways parabola needs

a \neq 0