Cleanup
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593527cb4e
commit
198f0a308d
@ -1,4 +1,5 @@
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#import "@preview/cetz:0.3.1"
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#import "handout.typ": *
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// Shorthand, we'll be using these a lot.
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#let tp = sym.plus.circle
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@ -35,3 +36,34 @@
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),
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)
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}
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/// Adds extra padding to an equation.
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/// Used as follows:
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///
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/// #eqmbox($
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/// f(x) = -2(x #tp 2)(x #tp 8)
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/// $)
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///
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/// Note that there are newlines between the $ and content,
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/// this gives us display math (which is what we want when using this macro)
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#let eqnbox(eqn) = {
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align(
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center,
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box(
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inset: 3mm,
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eqn,
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),
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)
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}
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#let dotline(a, b) = {
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cetz.draw.line(
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a,
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b,
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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}
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@ -72,36 +72,9 @@ Draw a graph of the tropical polynomial $f(x) = x^2 #tp 1x #tp 4$. \
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import cetz.draw: *
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let step = 0.75
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line(
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(0, 0),
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(4 * step, 8 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line(
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(0, 1 * step),
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(7 * step, 8 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line(
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(0, 4 * step),
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(8 * step, 4 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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dotline((0, 0), (4 * step, 8 * step))
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dotline((0, 1 * step), (7 * step, 8 * step))
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dotline((0, 4 * step), (8 * step, 4 * step))
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line((0, 0), (1 * step, 2 * step), (3 * step, 4 * step), (7.5 * step, 4 * step), stroke: 1mm + oblue)
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})
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@ -161,35 +134,9 @@ Graph $f(x) = -2x^2 #tp x #tp 8$. \
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import cetz.draw: *
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let step = 0.75
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line(
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(0, 0),
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(8 * step, 8 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line(
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(0.5 * step, 0),
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(4 * step, 8 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line(
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(0, 4 * step),
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(8 * step, 4 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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dotline((0, 0), (8 * step, 8 * step))
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dotline((0.5 * step, 0), (4 * step, 8 * step))
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dotline((0, 4 * step), (8 * step, 4 * step))
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line((0.5 * step, 0), (1 * step, 1 * step), (4 * step, 4 * step), (7.5 * step, 4 * step), stroke: 1mm + oblue)
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})
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@ -201,23 +148,15 @@ Find a factorization of $f$ in the form $a(x #tp r)(x#tp s)$.
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#solution([
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We (tropically) factor out $-2$ to get
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#align(
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center,
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box(
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inset: 3mm,
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$f(x) = -2(x^2 #tp 2x #tp 10)$,
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),
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)
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#eqnbox($
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f(x) = -2(x^2 #tp 2x #tp 10)
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$)
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by the same process as the previous problem, we get
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#align(
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center,
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box(
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inset: 3mm,
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$f(x) = -2(x #tp 2)(x #tp 8)$,
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),
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)
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#eqnbox($
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f(x) = -2(x #tp 2)(x #tp 8)
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$)
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])
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#v(1fr)
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@ -269,35 +208,9 @@ Graph $f(x) = 1x^2 #tp 3x #tp 5$.
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import cetz.draw: *
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let step = 0.75
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line(
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(0, 1 * step),
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(3.5 * step, 8 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line(
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(0, 4 * step),
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(8 * step, 4 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line(
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(0, 3 * step),
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(5 * step, 8 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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dotline((0, 1 * step), (3.5 * step, 8 * step))
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dotline((0, 5 * step), (8 * step, 5 * step))
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dotline((0, 3 * step), (5 * step, 8 * step))
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line((0, 1 * step), (2 * step, 5 * step), (7.5 * step, 5 * step), stroke: 1mm + oblue)
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})
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@ -308,13 +221,9 @@ Graph $f(x) = 1x^2 #tp 3x #tp 5$.
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Find a factorization of $f$ in the form $a(x #tp r)(x#tp s)$.
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#solution(
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align(
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center,
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box(
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inset: 3mm,
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$f(x) = 1x^2 #tp 3 x #tp 5 = 1(x #tp 2)^2$,
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),
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),
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eqnbox($
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f(x) = 1x^2 #tp 3 x #tp 5 = 1(x #tp 2)^2
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$),
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)
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#v(1fr)
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@ -343,35 +252,9 @@ Graph $f(x) = 2x^2 #tp 4x #tp 4$.
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import cetz.draw: *
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let step = 0.75
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line(
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(0, 2 * step),
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(3 * step, 8 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line(
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(0, 4 * step),
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(5 * step, 8 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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line(
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(0, 4 * step),
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(8 * step, 4 * step),
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stroke: (
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dash: "dashed",
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thickness: 0.5mm,
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paint: ored,
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),
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)
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dotline((0, 2 * step), (3 * step, 8 * step))
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dotline((0, 4 * step), (5 * step, 8 * step))
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dotline((0, 4 * step), (8 * step, 4 * step))
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line((0, 2 * step), (1 * step, 4 * step), (7.5 * step, 4 * step), stroke: 1mm + oblue)
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}),
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