High School Mathematics · Grades 10–12

10Quadratic Functions and Equations

Quadratic functions are the simplest functions after the linear ones, and the first whose graphs are genuinely curved. This chapter develops the complete toolbox for them: the canonical form, the discriminant, the sign of a quadratic expression, and the geometry of the parabola.

10.1 The canonical form

Definition 10.1 (Quadratic function)

A quadratic function is a function of the form

f(x) = ax^2 + bx + c, \qquad a \neq 0,

where $a$ , $b$ , $c$ are real numbers. The expression $ax^2 + bx + c$ is also called a quadratic trinomial.

Theorem 10.2 (Canonical form)

Every quadratic function can be written in exactly one way as

f(x) = a(x - \alpha)^2 + \beta, \qquad\text{where } \alpha = -\frac{b}{2a} \text{ and } \beta = f(\alpha).

This is the canonical form of $f$ .

Proof. We complete the square. Since $a \neq 0$ , factor it out of the first two terms:

ax^2 + bx + c = a\left(x^2 + \frac{b}{a}x\right) + c .

Now $x^2 + \frac{b}{a}x$ is the beginning of the square $\left(x + \frac{b}{2a}\right)^2 = x^2 + \frac ba x + \frac{b^2}{4a^2}$ , so $x^2 + \frac{b}{a}x = \left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}$ . Substituting:

f(x) = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c = a\bigl(x - \alpha\bigr)^2 + \beta,

with $\alpha = -\frac{b}{2a}$ and $\beta = c - \frac{b^2}{4a}$ . Evaluating at $x = \alpha$ kills the squared term, so $\beta = f(\alpha)$ . Uniqueness: expanding $a(x-\alpha)^2 + \beta$ and identifying the coefficients of $x$ forces $\alpha = -\frac{b}{2a}$ , and then $\beta$ is determined. ∎

Example 10.3

Let $f(x) = 2x^2 - 8x + 3$ . Then $\alpha = -\frac{-8}{2 \times 2} = 2$ and $\beta = f(2) = 8 - 16 + 3 = -5$ , so

f(x) = 2(x - 2)^2 - 5 .

Check by expanding: $2(x^2 - 4x + 4) - 5 = 2x^2 - 8x + 8 - 5 = 2x^2 - 8x + 3$ .

Proposition 10.4 (Extremum and variations)

Let $f(x) = a(x-\alpha)^2 + \beta$ .

If $a > 0$ , then $f$ is decreasing on $\intoc{-\infty}{\alpha}$ and increasing on $\intco{\alpha}{+\infty}$ : $f$ has a minimum $\beta$ , attained at $x = \alpha$ .
If $a < 0$ , the variations are reversed: $f$ has a maximum $\beta$ at $x = \alpha$ .

Proof. Suppose $a > 0$ and let $\alpha \leq u < v$ . Then

f(v) - f(u) = a\bigl((v-\alpha)^2 - (u-\alpha)^2\bigr) = a\,(v - u)\,\bigl((v-\alpha) + (u-\alpha)\bigr).

Each factor is positive: $a > 0$ , $v - u > 0$ , and $(v-\alpha) + (u-\alpha) > 0$ since $v > \alpha$ and $u \geq \alpha$ . So $f(v) > f(u)$ : $f$ is increasing on $\intco{\alpha}{+\infty}$ . For $u < v \leq \alpha$ the last factor is negative, so $f(v) < f(u)$ : $f$ is decreasing there. Finally $f(x) - f(\alpha) = a(x-\alpha)^2 \geq 0$ for all $x$ , so $\beta = f(\alpha)$ is the minimum. The case $a < 0$ follows by applying the above to $-f$ . ∎

Example 10.5

A farmer has $100$ meters of fence to enclose a rectangular field against a straight wall (no fence needed along the wall). If $x$ is the width, the enclosed area is $A(x) = x(100 - 2x) = -2x^2 + 100x$ . Here $\alpha = -\frac{100}{2\times(-2)} = 25$ and $A(25) = 25 \times 50 = 1250$ : the area is largest for a width of $25$ m, giving $1250$ m $^2$ .

10.2 Roots and factorization

Definition 10.6 (Root, discriminant)

A root of the trinomial $ax^2+bx+c$ is a solution of the equation $ax^2 + bx + c = 0$ . The discriminant of the trinomial is the number

\Delta = b^2 - 4ac .

Theorem 10.7 (Solving $ax^2+bx+c=0$ )

If $\Delta > 0$ , the equation has two distinct roots

$x_1 = \frac{-b - \sqrt{\Delta}}{2a}, \qquad x_2 = \frac{-b + \sqrt{\Delta}}{2a},$

and $ax^2+bx+c = a(x - x_1)(x - x_2)$ .
If $\Delta = 0$ , the equation has the single root $x_0 = -\frac{b}{2a}$ , and $ax^2+bx+c = a(x - x_0)^2$ .
If $\Delta < 0$ , the equation has no real solution, and the trinomial cannot be factored with real coefficients.

Proof. Start from the canonical form computed in Theorem 10.2:

ax^2+bx+c = a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a} = a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{\Delta}{4a^2}\right],

since $c - \frac{b^2}{4a} = \frac{4ac - b^2}{4a} = -\frac{\Delta}{4a} = a \times \left(-\frac{\Delta}{4a^2}\right)$ .

1. If $\Delta > 0$ , then $\frac{\Delta}{4a^2} = \left(\frac{\sqrt\Delta}{2a}\right)^2$ and the bracket is a difference of two squares:

ax^2+bx+c = a\left(x + \frac{b}{2a} - \frac{\sqrt\Delta}{2a}\right) \left(x + \frac{b}{2a} + \frac{\sqrt\Delta}{2a}\right) = a(x - x_2)(x - x_1).

A product is zero exactly when one of its factors is zero, giving the two roots, distinct because $\sqrt\Delta \neq 0$ .

2. If $\Delta = 0$ , the bracket is $\left(x + \frac{b}{2a}\right)^2$ , which vanishes only at $x_0 = -\frac{b}{2a}$ .

3. If $\Delta < 0$ , then $-\frac{\Delta}{4a^2} > 0$ , so the bracket is a square plus a positive number: it never vanishes, and the equation has no real solution. A factorization $a(x-x_1)(x-x_2)$ would produce roots, so none exists. ∎

Example 10.8

Solve $2x^2 - 8x + 3 = 0$ . Here $\Delta = (-8)^2 - 4 \times 2 \times 3 = 64 - 24 = 40 > 0$ , so there are two roots:

x_{1,2} = \frac{8 \pm \sqrt{40}}{4} = \frac{8 \pm 2\sqrt{10}}{4} = 2 \pm \frac{\sqrt{10}}{2}.

For $x^2 + x + 1 = 0$ : $\Delta = 1 - 4 = -3 < 0$ , no real solution.

Proposition 10.9 (Sum and product of the roots)

If $\Delta \geq 0$ and $x_1, x_2$ are the roots (equal when $\Delta = 0$ ), then

x_1 + x_2 = -\frac{b}{a}, \qquad x_1 x_2 = \frac{c}{a} .

Conversely, two numbers with sum $s$ and product $p$ are the solutions of $x^2 - sx + p = 0$ .

Proof. Expand the factorization of Theorem 10.7:

a(x - x_1)(x - x_2) = a x^2 - a(x_1 + x_2)x + a\,x_1 x_2 .

Identifying the coefficients with $ax^2 + bx + c$ gives $-a(x_1+x_2) = b$ and $a\,x_1x_2 = c$ . For the converse, if $u + v = s$ and $uv = p$ , then $(x-u)(x-v) = x^2 - sx + p$ , so $u$ and $v$ are its roots. ∎

Example 10.10

Find two numbers with sum $10$ and product $21$ . They solve $x^2 - 10x + 21 = 0$ ; $\Delta = 100 - 84 = 16$ , roots $\frac{10 \pm 4}{2} = 7$ and $3$ . Quick root spotting: since the coefficients of $x^2 - 5x + 4$ satisfy $1 - 5 + 4 = 0$ , the number $1$ is a root, and the other one is $\frac{c}{a} = 4$ (their product must be $\frac ca$ ).

10.3 Sign of a quadratic

Theorem 10.11 (Sign of a trinomial)

Let $f(x) = ax^2+bx+c$ with $a \neq 0$ .

If $\Delta > 0$ , with roots $x_1 < x_2$ : $f(x)$ has the sign of $a$ outside $\intcc{x_1}{x_2}$ , and the sign of $-a$ on $\intoo{x_1}{x_2}$ .
If $\Delta = 0$ : $f(x)$ has the sign of $a$ for all $x \neq x_0$ , and vanishes at $x_0$ .
If $\Delta < 0$ : $f(x)$ has the sign of $a$ for all $x$ .

In short: a trinomial has the sign of $a$ , except between its roots.

Proof. 1. Write $f(x) = a(x-x_1)(x-x_2)$ and follow the signs of the factors. If $x < x_1$ : both $x - x_1$ and $x - x_2$ are negative, their product is positive, so $f(x)$ has the sign of $a$ . If $x_1 < x < x_2$ : the factors have opposite signs, the product is negative, and $f(x)$ has the sign of $-a$ . If $x > x_2$ : both factors are positive and $f(x)$ has the sign of $a$ again.

2. $f(x) = a(x - x_0)^2$ and $(x-x_0)^2 > 0$ for $x \neq x_0$ .

3. From the proof of Theorem 10.7, $f(x) = a\left[\left(x + \frac{b}{2a}\right)^2 + \left(-\frac{\Delta}{4a^2}\right)\right]$ , and the bracket is always positive. ∎

Method 10.12 (Solving quadratic inequalities)

To solve $ax^2 + bx + c \leq 0$ (or $\geq$ , $<$ , $>$ ):

compute $\Delta$ and the roots, if any;
draw the sign table using Theorem 10.11 (sign of $a$ outside the roots);
read off the solution interval(s), paying attention to whether the roots themselves are included (wide inequality) or not (strict).

Example 10.13

Solve $-x^2 + 3x + 4 \geq 0$ . Here $\Delta = 9 + 16 = 25$ , and the roots are $\frac{-3 \pm 5}{-2}$ , that is $x_1 = -1$ and $x_2 = 4$ . Since $a = -1 < 0$ , the trinomial is negative outside the roots and positive between them. With the wide inequality, the solution set is $\intcc{-1}{4}$ .

10.4 The parabola

Definition 10.14 (Parabola)

The graph of a quadratic function $f(x) = a(x-\alpha)^2+\beta$ is a parabola with vertex $S(\alpha, \beta)$ . It opens upward if $a > 0$ , downward if $a < 0$ , and is symmetric about the vertical line $x = \alpha$ .

Remark 10.15

The symmetry is easy to check: for any $h$ , $f(\alpha + h) = ah^2 + \beta = f(\alpha - h)$ , so two points at equal horizontal distance from the line $x = \alpha$ have the same height.

The three positions of an upward parabola (a > 0) relative to the x-axis: two roots, one double root, no root. The vertex S has abscissa = -b/2a. — The three positions of an upward parabola ( $a > 0$ ) relative to the $x$ -axis: two roots, one double root, no root. The vertex $S$ has abscissa $\alpha = -\frac{b}{2a}$ .

Sign of a downward trinomial (a < 0) with two roots: the sign of -a between the roots (shaded), the sign of a outside. — Sign of a downward trinomial ( $a < 0$ ) with two roots: the sign of $-a$ between the roots (shaded), the sign of $a$ outside.

Method 10.16 (Reading a parabola)

To sketch $f(x) = ax^2+bx+c$ : find the orientation (sign of $a$ ), the vertex $\left(-\frac{b}{2a},\, f\!\left(-\frac{b}{2a}\right)\right)$ , the intersection $(0, c)$ with the $y$ -axis, and the roots, if any. The axis of symmetry $x = -\frac{b}{2a}$ then places every computed point twice.

Example 10.17

For $f(x) = x^2 - 4x + 3$ : upward, vertex $(2, -1)$ , crosses the $y$ -axis at $(0,3)$ , roots $1$ and $3$ (spotted: $1 - 4 + 3 = 0$ ). By symmetry, the point $(4, 3)$ is also on the curve, mirroring $(0, 3)$ .

10.5 Exercises

Exercise 10.1 ★

Solve the equations

x^2 - 5x + 6 = 0, \qquad 2x^2 + 3x - 2 = 0, \qquad x^2 + 2x + 5 = 0, \qquad 9x^2 - 6x + 1 = 0 .

Solution

Solution of Exercise 10.1.

$x^2-5x+6$ : $\Delta = 25 - 24 = 1$ , roots $\frac{5 \pm 1}{2}$ , i.e. $2$ and $3$ .

$2x^2+3x-2$ : $\Delta = 9 + 16 = 25$ , roots $\frac{-3 \pm 5}{4}$ , i.e. $\frac12$ and $-2$ .

$x^2+2x+5$ : $\Delta = 4 - 20 = -16 < 0$ , no real solution.

$9x^2-6x+1$ : $\Delta = 36 - 36 = 0$ , one (double) root $x_0 = \frac{6}{18} = \frac13$ .

Exercise 10.2 ★

Write in canonical form and give the vertex of the parabola:

f(x) = x^2 - 6x + 11, \qquad g(x) = -2x^2 + 4x + 1, \qquad h(x) = 3x^2 + 6x .

Solution

Solution of Exercise 10.2.

$f(x) = x^2 - 6x + 11 = (x-3)^2 - 9 + 11 = (x-3)^2 + 2$ : vertex $(3, 2)$ .

$g(x) = -2x^2+4x+1 = -2(x^2 - 2x) + 1 = -2\bigl((x-1)^2 - 1\bigr) + 1 = -2(x-1)^2 + 3$ : vertex $(1, 3)$ .

$h(x) = 3x^2 + 6x = 3(x^2 + 2x) = 3\bigl((x+1)^2 - 1\bigr) = 3(x+1)^2 - 3$ : vertex $(-1, -3)$ .

Exercise 10.3 ★

Solve the inequalities

x^2 - 3x - 10 < 0, \qquad 2x^2 + x + 3 > 0, \qquad -x^2 + 6x - 9 \geq 0 .

Solution

Solution of Exercise 10.3.

$x^2 - 3x - 10$ : $\Delta = 9 + 40 = 49$ , roots $\frac{3 \pm 7}{2} = -2$ and $5$ ; positive leading coefficient, so the trinomial is negative strictly between the roots: the solution set is $\intoo{-2}{5}$ .

$2x^2 + x + 3$ : $\Delta = 1 - 24 = -23 < 0$ and $a = 2 > 0$ : always positive, the solution set is $\R$ .

$-x^2 + 6x - 9 = -(x-3)^2$ : it is $\geq 0$ only where $(x-3)^2 \leq 0$ , that is at $x = 3$ . The solution set is $\{3\}$ .

Exercise 10.4 ★

Without computing $\Delta$ , find one obvious root, then the other:

x^2 - 7x + 6 = 0, \qquad 2x^2 + 5x - 7 = 0, \qquad x^2 + 4x - 5 = 0 .

(Hint: test $x = 1$ and $x = -1$ , then use the product of the roots.)

Solution

Solution of Exercise 10.4.

$x^2 - 7x + 6$ : $1 - 7 + 6 = 0$ , so $1$ is a root; the product of the roots is $6$ , so the other root is $6$ .

$2x^2 + 5x - 7$ : $2 + 5 - 7 = 0$ , so $1$ is a root; the product is $\frac{c}{a} = -\frac72$ , so the other root is $-\frac72$ .

$x^2 + 4x - 5$ : $1 + 4 - 5 = 0$ , so $1$ is a root; the product is $-5$ , so the other root is $-5$ .

Exercise 10.5 ★★

Find two numbers whose sum is $14$ and whose product is $45$ . Then find the dimensions of a rectangle of perimeter $34$ and area $60$ .

Solution

Solution of Exercise 10.5.

The two numbers solve $x^2 - 14x + 45 = 0$ (Proposition 10.9); $\Delta = 196 - 180 = 16$ , roots $\frac{14 \pm 4}{2} = 9$ and $5$ .

For the rectangle: half the perimeter gives length $+$ width $= 17$ , and the area gives their product $60$ . They solve $x^2 - 17x + 60 = 0$ ; $\Delta = 289 - 240 = 49$ , roots $\frac{17 \pm 7}{2} = 12$ and $5$ . The rectangle is $12 \times 5$ .

Exercise 10.6 ★★

Let $f(x) = x^2 + (m-2)x + 1$ , where $m$ is a real parameter. For which values of $m$ does the equation $f(x) = 0$ have exactly one solution? No solution?

Solution

Solution of Exercise 10.6.

$\Delta = (m-2)^2 - 4$ . Exactly one solution when $\Delta = 0$ : $(m-2)^2 = 4$ , so $m - 2 = \pm 2$ , i.e. $m = 0$ or $m = 4$ . No solution when $\Delta < 0$ : $(m-2)^2 < 4$ , i.e. $-2 < m - 2 < 2$ , that is $0 < m < 4$ .

Exercise 10.7 ★★

A ball is thrown upward; its height after $t$ seconds is $h(t) = -5t^2 + 20t + 1$ (in meters). What is the maximal height reached, and at what time? During which time interval is the ball at least $16$ m high?

Solution

Solution of Exercise 10.7.

The vertex is at $t = -\frac{20}{2 \times (-5)} = 2$ , and $h(2) = -20 + 40 + 1 = 21$ : maximal height $21$ m, reached after $2$ seconds.

$h(t) \geq 16$ means $-5t^2 + 20t - 15 \geq 0$ , i.e. (dividing by $-5$ , which reverses the inequality) $t^2 - 4t + 3 \leq 0$ . The roots of $t^2-4t+3$ are $1$ and $3$ ( $1 - 4 + 3 = 0$ , product $3$ ), so $t^2 - 4t + 3 \leq 0$ exactly on $\intcc{1}{3}$ : the ball is at least $16$ m high between the times $t = 1$ s and $t = 3$ s.

Exercise 10.8 ★★

Solve the equation $x^4 - 13x^2 + 36 = 0$ by setting $X = x^2$ (a biquadratic equation).

Solution

Solution of Exercise 10.8.

With $X = x^2$ : $X^2 - 13X + 36 = 0$ , $\Delta = 169 - 144 = 25$ , so $X = \frac{13 \pm 5}{2} = 9$ or $4$ . Back to $x$ : $x^2 = 9$ gives $x = \pm 3$ , and $x^2 = 4$ gives $x = \pm 2$ . The solution set is $\{-3, -2, 2, 3\}$ .

Exercise 10.9 ★★

For which values of $x$ is the point $M(x, x^2)$ of the parabola $y = x^2$ strictly below the line $y = x + 2$ ? Interpret on a sketch.

Solution

Solution of Exercise 10.9.

The condition is $x^2 < x + 2$ , i.e. $x^2 - x - 2 < 0$ . Since $x^2 - x - 2 = (x+1)(x-2)$ (roots $-1$ and $2$ ), the inequality holds strictly between the roots: $x \in \intoo{-1}{2}$ . On a sketch, the parabola $y = x^2$ dips below the line $y = x+2$ exactly between their two intersection points $(-1, 1)$ and $(2, 4)$ .

Exercise 10.10 ★★

The sum of a positive number and its reciprocal is $\frac{13}{6}$ . Find the number. (Reduce to a quadratic equation.)

Solution

Solution of Exercise 10.10.

Let $x > 0$ satisfy $x + \frac1x = \frac{13}{6}$ . Multiplying by $6x > 0$ : $6x^2 + 6 = 13x$ , i.e. $6x^2 - 13x + 6 = 0$ . Then $\Delta = 169 - 144 = 25$ and $x = \frac{13 \pm 5}{12}$ , giving $x = \frac32$ or $x = \frac23$ . Both are positive and are reciprocals of each other, as expected: the number is $\frac32$ (or $\frac23$ ).

Exercise 10.11 ★★★

Let $\mathcal P$ be the parabola $y = x^2$ and let $d_m$ be the line $y = mx - 1$ , where $m$ is a real parameter.

For which values of $m$ do $\mathcal P$ and $d_m$ intersect in two points? One point? No point?
When they touch in exactly one point, compute the contact point. (In Chapter 12 you will recognize $d_m$ as the tangent to $\mathcal P$ at that point.)

Solution

Solution of Exercise 10.11.

1. Intersections correspond to solutions of $x^2 = mx - 1$ , i.e. $x^2 - mx + 1 = 0$ , whose discriminant is $\Delta = m^2 - 4$ . Two points when $\abs{m} > 2$ ; exactly one when $m = \pm 2$ ; none when $-2 < m < 2$ .

2. For $m = 2$ : $x^2 - 2x + 1 = (x-1)^2 = 0$ , so the unique contact point is $(1, 1)$ . For $m = -2$ : $(x+1)^2 = 0$ , contact point $(-1, 1)$ . In both cases the line meets the parabola at a single point without crossing it — it is the tangent there, and indeed its slope $m = \pm 2$ equals the value at $x = \pm 1$ of the derivative $2x$ of $x^2$ (Chapter 12).

10.6 Problem: The parabola’s secret point

Problem 10.1

Weekend problem — focus and directrix: why satellite dishes, headlights and suspension bridges all draw the same curve, of which there is essentially only one

Every satellite dish on every balcony aims its receiver at one precise interior point; every headlight hides its bulb at the same special spot. That point — the focus — is a secret the parabola has kept since Greek geometry, and the distance formula is enough to extract it. Along the way, this problem drills the discriminant, throws projectiles over walls, suspends a bridge, and proves an astonishing final fact: up to zoom, there is only one parabola in the world.

Part I — Trinomial fluency.

Solve $x^2 - 5x + 6 = 0$ , then $2x^2 + x - 6 = 0$ , then $x^2 + x + 1 = 0$ (Theorem 10.7).
Write $2x^2 - 4x - 6$ in factored form and in canonical form (Theorem 10.2). Together with the expanded form, you now hold three costumes of one function: say which feature each displays at a glance (roots, vertex, $y$ -intercept).
Solve $-x^2 + 4x - 3 \geq 0$ with the sign rule (Theorem 10.11).
Viète as detective (Proposition 10.9, continuing Babylon in Problem 2.1): find two numbers with sum $7$ and product $12$ by writing their equation. Then: one root of $x^2 - 5x + c = 0$ equals $2$ ; find $c$ and the other root without solving anything.
For which values of $m$ does $x^2 + mx + 9 = 0$ have a double root? No root? Two roots?

Part II — The focus unveiled. Work on the parabola $y = x^2$ , the point $F\left(0, \frac14\right)$ and the horizontal line $\delta$ of equation $y = -\frac14$ .

For a point $P(x, x^2)$ of the parabola, expand $PF^2 = x^2 + \left(x^2 - \frac14\right)^2$ and show it is the perfect square $\left(x^2 + \frac14\right)^2$ . Deduce $PF$ .
Compute the distance from $P$ to the line $\delta$ , and conclude: every point of the parabola is exactly as far from $F$ as from $\delta$ . ( $F$ is the focus, $\delta$ the directrix.)
Prove the converse: a point $P(x, y)$ with $PF = \operatorname{dist}(P, \delta)$ must satisfy $y = x^2$ . (Square both distances and simplify.) The parabola is thus a locus, exactly as the circle was in Problem 5.1: equidistance from a point and a line, instead of from one point.
For the wider parabola $y = \frac{x^2}{4f}$ (with $f > 0$ ), adapt question 6 to show the focus is $(0, f)$ and the directrix $y = -f$ . Application: a satellite dish is $60$ cm wide and $9$ cm deep — its rim passes through $(30, 9)$ . Find $f$ : how high above the vertex must the receiver hang?
The reason dishes work: rays arriving parallel to the axis all reflect through the focus (and a bulb at the focus throws a parallel beam — headlights are dishes run backwards). Explain why question 7’s equidistance makes this plausible (think of the rays as marching wavefronts), and say which tool, coming in Chapter 12 (see Exercise 10.11), turns plausibility into proof.

Part III — Parabolas at work.

A ball follows the trajectory $y = x - \frac{x^2}{20}$ ( $x$ and $y$ in meters). Find where it lands, the horizontal position of its highest point, and the maximal height (Proposition 10.4).
On the same trajectory, a wall stands at $x = 4$ . Compute the ball’s height there: does it clear a $3$ m wall? A $4$ m wall?
A parabolic arch spans $40$ m with maximal height $10$ m: in centered coordinates, $y = 10 - \frac{x^2}{40}$ . Can a truck $7$ m tall pass at $10$ m from the center? At $15$ m?
A theater’s revenue at ticket price $p$ euros is $R(p) = p\,(120 - 2p)$ (demand falls as price rises). Find the price maximizing revenue, and the maximum.
The Golden Gate’s main cables are parabolic to excellent accuracy (a uniform deck load produces a parabola — a freely hanging chain does not: its curve, the catenary, needs the exponential of grade 12). With span $1\,280$ m and sag $143$ m, the cable follows $y = 143\left(\frac{x}{640}\right)^2$ from the center. How high is the cable above its lowest point at $x = 320$ m?

Part IV — Lines, tangents, and the last surprise.

Complete Exercise 10.11: the lines $y = mx - 1$ through the external point $(0, -1)$ meet $y = x^2$ according to the sign of a discriminant. Find the two tangent lines from $(0, -1)$ and their contact points.
Now the lines $y = mx + 1$ through the interior point $(0, 1)$ : compute the discriminant of the intersection equation and show every such line cuts the parabola twice. Moral: from inside, no tangents — just as for a circle.
The focal chord: intersect $y = x^2$ with the horizontal line through the focus, $y = \frac14$ . What is the chord’s length? Check the general rule (worth remembering for dishes): for $y = \frac{x^2}{4f}$ , the chord through the focus parallel to the directrix has length $4f$ .
The last surprise: apply the zoom $(x, y) \mapsto (kx, ky)$ to the parabola $y = x^2$ and show the image is exactly $y = \frac{x^2}{k}$ . Conclude: every parabola $y = \frac{x^2}{4f}$ is a magnified unit parabola — unlike circles and ellipses of different proportions, all parabolas are similar.
Finale: the parabola’s portrait in five lines — an equation with three costumes (question 2), a locus (questions 7–8), an engineer’s reflector (questions 9, 10), a projectile’s path and a bridge’s cable (questions 11–15), and a single curve up to zoom (question 19). One sentence each.

Solution

Solution of Problem 10.1.

1. $x^2 - 5x + 6$ : $\Delta = 1$ , roots $2$ and $3$ . $2x^2 + x - 6$ : $\Delta = 1 + 48 = 49$ , roots $\frac{-1 \pm 7}{4}$ : $\frac32$ and $-2$ . $x^2 + x + 1$ : $\Delta = -3 < 0$ : no real solution.

2. $\Delta = 16 + 48 = 64$ : roots $3$ and $-1$ , so $2x^2 - 4x - 6 = 2(x - 3)(x + 1)$ ; canonical form $2(x - 1)^2 - 8$ . The factored form shows the roots, the canonical form the vertex $(1, -8)$ , the expanded form the $y$ -intercept $-6$ .

3. Roots $1$ and $3$ , leading coefficient negative: the trinomial is $\geq 0$ between the roots: $x \in \intcc{1}{3}$ .

4. The numbers are the roots of $x^2 - 7x + 12 = 0$ : $3$ and $4$ . If $2$ is a root of $x^2 - 5x + c$ , the other root $r$ satisfies $2 + r = 5$ (sum), so $r = 3$ , and $c = 2 \times 3 = 6$ (product): Viète reads it off.

5. $\Delta = m^2 - 36$ : double root for $m = \pm 6$ (root $\mp 3$ ); no root for $-6 < m < 6$ ; two roots for $\abs m > 6$ .

6. $PF^2 = x^2 + \left(x^2 - \frac14\right)^2 = x^2 + x^4 - \frac{x^2}{2} + \frac{1}{16} = x^4 + \frac{x^2}{2} + \frac{1}{16} = \left(x^2 + \frac14\right)^2$ . Hence $PF = x^2 + \frac14$ (both factors positive).

7. The distance from $P(x, x^2)$ to the horizontal line $y = -\frac14$ is $x^2 + \frac14$ — exactly $PF$ . Every point of the parabola is equidistant from the focus and the directrix.

8. $PF^2 = x^2 + \left(y - \frac14\right)^2$ and $\operatorname{dist}^2 = \left(y + \frac14\right)^2$ . Equality gives $x^2 + y^2 - \frac y2 + \frac{1}{16} = y^2 + \frac y2 + \frac1{16}$ , so $x^2 = y$ : the locus is exactly the parabola. One point and one line generate the curve, as one point and one distance generated the circle.

9. For $y = \frac{x^2}{4f}$ , the same expansion with $F(0, f)$ gives $PF^2 = x^2 + \left(\frac{x^2}{4f} - f\right)^2 = \left(\frac{x^2}{4f} + f\right)^2$ : distance to $F$ equals distance to $y = -f$ . Dish: $9 = \frac{30^2}{4f}$ gives $4f = 100$ , $f = 25$ cm: the receiver hangs a quarter meter above the vertex — outside the $9$ cm bowl, as real dishes show.

10. A wavefront arriving parallel to the axis has travelled equal distances to reach the directrix’s level; equidistance converts “equal distance to the directrix” into “equal distance to $F$ ”: all the reflected signals arrive at $F$ in step, reinforcing — that is where to listen. The honest proof needs the tangent’s slope at each point — the derivative of Chapter 12, glimpsed in Exercise 10.11.

11. $y = x\left(1 - \frac{x}{20}\right)$ : lands at $x = 20$ m. Vertex midway: $x = 10$ , height $y(10) = 10 - 5 = 5$ m.

12. $y(4) = 4 - \frac{16}{20} = 3.2$ m: clears $3$ m with room, smacks into a $4$ m wall.

13. $y(10) = 10 - \frac{100}{40} = 7.5$ m $> 7$ : the truck passes at $10$ m off-center. $y(15) = 10 - \frac{225}{40} = 4.375$ m: not at $15$ m.

14. $R(p) = -2p^2 + 120p$ : vertex at $p = \frac{120}{4} = 30$ euros, revenue $R(30) = 30 \times 60 = 1\,800$ euros.

15. $y(320) = 143 \times \left(\frac{320}{640}\right)^2 = \frac{143}{4} \approx 36$ m above the lowest point.

16. $x^2 = mx - 1$ gives $x^2 - mx + 1 = 0$ , $\Delta = m^2 - 4$ : two points for $\abs m > 2$ , none for $\abs m < 2$ , tangency for $m = \pm 2$ with contact $x = \frac m2 = \pm 1$ : tangents $y = 2x - 1$ at $(1, 1)$ and $y = -2x - 1$ at $(-1, 1)$ .

17. $x^2 = mx + 1$ gives $\Delta = m^2 + 4 > 0$ for every $m$ : two intersections always. From an interior point, every line crosses the curve twice — tangents are drawn only from outside, exactly as for circles.

18. $y = \frac14$ meets $y = x^2$ at $x = \pm\frac12$ : chord length $1$ . For $y = \frac{x^2}{4f}$ : $y = f$ gives $x^2 = 4f^2$ , $x = \pm 2f$ : length $4f$ — the focal width is four times the focal distance (dish designers call it the latus rectum).

19. A point of $y = x^2$ is $(a, a^2)$ ; its image is $(X, Y) = (ka, ka^2)$ , and $\frac{X^2}{k} = \frac{k^2a^2}{k} = ka^2 = Y$ : the image curve is exactly $y = \frac{x^2}{k}$ . Choosing $k = 4f$ reaches every parabola: they are all magnifications of one master curve. (Circles are all similar too — but ellipses are not: squash matters, zoom does not.)

20. An equation with three readable costumes; the locus of equidistance from a point and a line; the reflector that gathers parallel rays at its focus and the cable that carries a uniform deck; the path of every thrown ball; and, up to magnification, a single curve — one parabola, everywhere.