Mathematics for
Data Science 1

1.1Reading a, b and c

Example 1: identify a, b and c

Write each function in the form $ax^2 + bx + c$ and read off a, b, c.

Take the sign with the number: in (iii) the coefficient of $x^2$ is $-1$, and in (iv) $b = 0$ because there is no x term. The functions (i) and (ii) are the ones these notes use for all examples; (iii) and (iv) are extra illustrations.

1.2Evaluating the function

Example 2: values of f(x) = x² + 2x − 24

Find $f(0)$, $f(4)$, $f(-6)$ and $f(-1)$.

These four points $(0, -24)$, $(4, 0)$, $(-6, 0)$ and $(-1, -25)$ are the y-intercept, the two x-intercepts and the vertex of Figure 1. Note that $f(0) = c$ always, so $c$ is where the curve meets the y-axis.

1.3Going deeper

1. Three numbers fix the parabola. Change $a$ and the curve gets wider, narrower or flips; change $c$ and it slides up or down; $b$ moves it sideways and up or down together (extra remark).
2. Always one y-intercept. $f(0) = c$, so a quadratic function meets the y-axis exactly once, at $(0, c)$.
3. Name the type first. If a problem gives $x^2 + 2x = 24$, move everything to one side to see $a$, $b$, $c$ before doing anything else.

1. x-coordinate of the vertex. Compute $x = \frac{-b}{2a}$.
2. y-coordinate of the vertex. Put this $x$ into the function: $y = f\left(\frac{-b}{2a}\right)$.

2.1Worked examples

Example 3: vertex of x² + 2x − 24

1. Here $a = 1$, $b = 2$, $c = -24$.
2. Axis. $x = \frac{-2}{2(1)} = -1$.
3. y-coordinate. $f(-1) = (-1)^2 + 2(-1) - 24 = 1 - 2 - 24 = -25$.

Example 4: vertex of 4x² + 9x − 9

1. Here $a = 4$, $b = 9$, $c = -9$.
2. Axis. $x = \frac{-9}{2(4)} = -\frac{9}{8}$.
3. y-coordinate. $f\left(-\frac{9}{8}\right) = 4 \cdot \frac{81}{64} - \frac{81}{8} - 9 = \frac{81}{16} - \frac{162}{16} - \frac{144}{16} = -\frac{225}{16}$.

2.2Why folding works: mirror pairs

2.3Why the formula works

Completing the square (Chapter 7) gives

The bracket squared is never negative and is zero exactly when $x = \frac{-b}{2a}$. So the curve is symmetric about that line, and the vertex has height $c - \frac{b^2}{4a}$, which agrees with $f\left(\frac{-b}{2a}\right)$. The lecture states the axis and vertex formulas without proof; this short derivation is added to show where they come from.

2.4Going deeper

1. The axis is a vertical line, so it is written $x = \ldots$; the vertex is a point, written as a pair.
2. Quick check. The vertex of $x^2 + 2x - 24$ is the midpoint of its x-intercepts $-6$ and $4$: $\frac{-6 + 4}{2} = -1$ (see Chapter 8).
3. Fractions are normal. The vertex of $4x^2 + 9x - 9$ is $\left(-\frac{9}{8}, -\frac{225}{16}\right)$; keep it as a fraction and convert to a decimal only at the end.

The sign of $a$ decides which way the parabola opens.

3.1Minimum and maximum values

Example 5: minimum or maximum?

For each function decide whether it has a minimum or a maximum, and find its value and where it occurs: (i) $x^2 + 2x - 24$, (ii) $4x^2 + 9x - 9$, (iii) $-x^2 - 2x + 24$.

• (i) $a = 1 > 0$, so a minimum. It occurs at $x = -1$ and equals $f(-1) = -25$.
• (ii) $a = 4 > 0$, so a minimum. It occurs at $x = -\frac{9}{8}$ and equals $-\frac{225}{16} = -14.0625$.
• (iii) $a = -1 < 0$, so a maximum. The axis is $x = \frac{-(-2)}{2(-1)} = -1$ and the maximum value is $f(-1) = -1 + 2 + 24 = 25$.

So (i) never goes below $-25$, (ii) never goes below $-14.0625$, and (iii) never goes above $25$. These three computations apply the rule with the vertex formula.

3.2Going deeper

1. Range of the function. For $a > 0$ the values of $f$ are all numbers from the minimum upwards; for $a < 0$, all numbers from the maximum downwards. For (i) that is $f(x) \ge -25$ (extra remark).
2. Optimisation. Questions like “the least cost” or “the greatest height” of a quadratic are asking for the vertex value.
3. Minimum value is a y-value. Say where it occurs ($x$) and what it is ($y$) separately.

A straight line has one slope everywhere. A parabola bends, so its slope changes from point to point; the formula $2ax + b$ gives the slope at each $x$ (the slope of the tangent line touching the curve there).

4.1Worked examples

Example 6: slope of x² + 2x − 24

1. $a = 1$, $b = 2$, so the slope at $x$ is $2(1)x + 2 = 2x + 2$.
2. Substitute:

Example 7: slope of 4x² + 9x − 9

Find the slope at $x = -3, -\frac{9}{8}, 0$ and $\frac{3}{4}$.

1. $a = 4$, $b = 9$, so the slope is $8x + 9$.
2. Substitute:

4.2What the slope tells us

The lecture gives only the formula $2ax + b$. The two observations above are direct consequences of it, using the vertex formula of Chapter 2.

4.3Going deeper

1. A straight line is the special case. With $a = 0$ the formula $2ax + b$ becomes just $b$, the constant slope of $bx + c$ from Week 2.
2. The slope changes by $2a$ per unit. In $2x + 2$ it grows by 2 each step: $-8, -6, \ldots$. The bigger $|a|$, the faster the curve bends (extra remark).
3. Use it to find the vertex. Set $2ax + b = 0$ and solve: you get $x = \frac{-b}{2a}$ again.

If a quadratic function is set equal to a value, the result is a quadratic equation.

So one method for finding the roots of $ax^2 + bx + c = 0$ is to find the zeros of the quadratic function $f(x) = ax^2 + bx + c$.

5.1Putting an equation in standard form

Example 8: standard form

Write (i) $x^2 + 2x = 24$ and (ii) $4x^2 + 9x = 9$ in standard form and read off a, b, c.

• (i) Subtract 24 from both sides: $x^2 + 2x - 24 = 0$, so $a = 1$, $b = 2$, $c = -24$.
• (ii) Subtract 9 from both sides: $4x^2 + 9x - 9 = 0$, so $a = 4$, $b = 9$, $c = -9$.

5.2Roots and the graph

The equation $x^2 + 2x = 24$ says: the function $x^2 + 2x$ has value 24. Its solutions are the x-values where the curve is at height 24, that is, where the horizontal line $y = 24$ meets it. Moving 24 to the left gives the same solutions as the x-intercepts of $y = x^2 + 2x - 24$.

5.3Going deeper

1. Root, zero, x-intercept: one idea. For $x^2 + 2x - 24$ the numbers $-6$ and $4$ are the roots of the equation, the zeros of the function and the x-coordinates of the intercepts.
2. Why standard form first. Every method of the next chapters starts from $ax^2 + bx + c = 0$; a root only appears when one side is zero.
3. Sum of the roots (extra remark). $-6 + 4 = -2 = -\frac{b}{a}$ and $-3 + \frac{3}{4} = -\frac{9}{4} = -\frac{b}{a}$: the roots add up to $-\frac{b}{a}$, twice the axis.

6.1Steps

1. Write the equation in standard form (equal to 0).
2. Factor the polynomial.
3. Use the Zero Product Property to set each factor equal to zero.
4. Solve each resulting linear equation.

6.2Lecture examples

Example 9: solve x² + 2x = 24 by factoring

1. Standard form. $x^2 + 2x - 24 = 0$.
2. Factor. Split $2x$ as $6x - 4x$:
3. Zero Product Property. $x - 4 = 0$ or $x + 6 = 0$.
4. Solve. $x = 4$ or $x = -6$.

Check. $4^2 + 2(4) = 24$ and $(-6)^2 + 2(-6) = 36 - 12 = 24$. Both work.

Example 10: solve 4x² + 9x = 9 by factoring

1. Standard form. $4x^2 + 9x - 9 = 0$.
2. Factor. Split $9x$ as $12x - 3x$:
3. Zero Product Property. $4x - 3 = 0$ or $x + 3 = 0$.
4. Solve. $x = \frac{3}{4}$ or $x = -3$.

Check. $4\left(\frac{3}{4}\right)^2 + 9 \cdot \frac{3}{4} = \frac{9}{4} + \frac{27}{4} = 9$ and $4(9) + 9(-3) = 36 - 27 = 9$.

6.3Going deeper

1. Why the right side must be 0. The Zero Product Property says nothing about a product equal to 24: $(x-1)(x+3) = 24$ does not mean a factor is 24 or 0. Always move everything to one side first.
2. Finding the split. For $x^2 + 2x - 24$, list pairs with product $-24$: $1, -24$; $-1, 24$; $2, -12$; $-2, 12$; $3, -8$; $-3, 8$; $4, -6$; $-4, 6$. Only $-4, 6$ adds to 2.
3. Always check by expanding. $(x-4)(x+6) = x^2 + 6x - 4x - 24 = x^2 + 2x - 24$. If this does not return the original, the factoring is wrong.
4. When factoring is hard. It works when the polynomial can be factored into nice integers. When it cannot, use completing the square or the quadratic formula.

7.1Steps

1. Transform the equation so that a perfect square is on one side and a constant is on the other.
2. Take the square root of each side. Remember that the square root of a constant gives a positive and a negative value.
3. Solve each resulting equation. (If you are taking the square root of a negative number, there is no real solution.)

7.2Lecture examples

Example 11: solve x² + 2x = 24 by completing the square

1. Add and subtract 1 (half of 2, squared): $x^2 + 2x + 1 - 1 = 24$.
2. Perfect square. $(x + 1)^2 = 25 \;\Rightarrow\; x + 1 = \pm\sqrt{25}$.
3. $x + 1 = +5$ or $x + 1 = -5$.
4. $x = 4$ or $x = -6$.

Example 12: solve 4x² + 9x = 9 by completing the square

1. Write $4x^2$ as $(2x)^2$ and $9x$ as $2 \cdot 2x \cdot \frac{9}{4}$, then add and subtract $\left(\frac{9}{4}\right)^2 = \frac{81}{16}$:
2. The first three terms are a perfect square, and $9 + \frac{81}{16} = \frac{225}{16}$:
3. $2x + \frac{9}{4} = \frac{15}{4}$ or $2x + \frac{9}{4} = -\frac{15}{4}$.
4. $2x = \frac{6}{4} \Rightarrow x = \frac{3}{4}$ or $2x = -\frac{24}{4} = -6 \Rightarrow x = -3$.

7.3When there is no real solution

If the right-hand side after completing the square is negative, there is no real solution because a real square is never negative. For example $x^2 + 2x + 5 = 0$ becomes $(x + 1)^2 = -4$, which has no real solution. The slides state the rule without an example; this equation is an illustration.

7.4Going deeper: the general recipe

1. Make the coefficient of $x^2$ equal to 1 (or write $ax^2$ as a square, as in Example 12).
2. Half the coefficient of $x$, squared. For $x^2 + 2x$ it is $\left(\frac{2}{2}\right)^2 = 1$. Add and subtract it.
3. The same idea finds the vertex. $x^2 + 2x - 24 = (x + 1)^2 - 25$, so the vertex is $(-1, -25)$ (extra remark).
4. Do not forget the $\pm$. Dropping the negative root loses half the answer.

The expression $b^2 - 4ac$ is called the discriminant.

8.1Number of real roots

8.2Lecture examples

Example 13: solve x² + 2x − 24 = 0 by the formula

1. $a = 1$, $b = 2$, $c = -24$.
2. Discriminant. $b^2 - 4ac = 2^2 - 4(1)(-24) = 4 + 96 = 100 > 0$, so two real roots.
3. Substitute. $x = \frac{-2 \pm \sqrt{100}}{2} = \frac{-2 \pm 10}{2}$.
4. $x = \frac{-2 + 10}{2} = 4$ or $x = \frac{-2 - 10}{2} = -6$.

Example 14: solve 4x² + 9x − 9 = 0 by the formula

1. $a = 4$, $b = 9$, $c = -9$.
2. Discriminant. $b^2 - 4ac = 9^2 - 4(4)(-9) = 81 + 144 = 225 > 0$, so two real roots.
3. Substitute. $x = \frac{-9 \pm \sqrt{225}}{8} = \frac{-9 \pm 15}{8}$.
4. $x = \frac{-9 + 15}{8} = \frac{3}{4}$ or $x = \frac{-9 - 15}{8} = -3$.

8.3The same equations by all three methods

8.4Check with the vertex

The axis of symmetry lies halfway between the two roots. For the first equation, $\frac{4 + (-6)}{2} = -1$, the axis found in Chapter 2. For the second, $\frac{\frac{3}{4} + (-3)}{2} = -\frac{9}{8}$, again the axis. The formula shows why: the two roots are $\frac{-b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$, equally spaced about $\frac{-b}{2a}$. At each root the slope $2ax + b$ equals $\pm\sqrt{b^2 - 4ac}$: $\pm 10$ for the first equation and $\pm 15$ for the second. This check follows from the formulas; it is not on the slides.

8.5Where the formula comes from

Start from $ax^2 + bx + c = 0$, complete the square as in Chapter 2, and take square roots:

This also explains the table: the square root is real only when $b^2 - 4ac \ge 0$, and it vanishes when $b^2 - 4ac = 0$, leaving a single root. This derivation is added to show where the formula and the discriminant table come from; the slides state them directly.

8.6Going deeper

1. Compute the discriminant first. It tells you whether to expect real roots before you take any square root. A perfect-square discriminant (100, 225) means the roots are rational and the equation factors.
2. Which method? Factoring is fastest when the numbers are nice; completing the square shows the vertex; the formula always works.
3. Parentheses with negative $b$ and $c$. Write $-b$ as $-(2) = -2$ and $4ac$ with its signs: $4(1)(-24) = -96$, so $b^2 - 4ac = 4 + 96$.

Everything in Week 3 on one sheet, for a last-minute revision.

Formulas

Shape and roots at a glance

Three methods

Values computed in the notes