Quadratic Functions and Parabolas: Free Response

Expand the square first, then distribute the $\tfrac{1}{2}$ across every term it produced.

$g(x) = \tfrac{1}{2}(x^2 - 16x + 64) + 6$

$= \tfrac{1}{2}x^2 - 8x + 32 + 6 = \tfrac{1}{2}x^2 - 8x + 38$

Matching the template $ax^2 + bx + c$ gives $a = \tfrac{1}{2}$, $b = -8$ and $c = 38$. The leading coefficient came through the change of costume untouched, as it always does, since expanding $a(x - h)^2 + k$ can only start $ax^2 + \cdots$.

Each form is silent about what the other announces. Vertex form matches $a(x - h)^2 + k$ with $h = 8$ and $k = 6$, so the vertex $(8, 6)$, the axis $x = 8$ and the range are readable with no computation at all. Standard form gives $g(0) = c = 38$, the height at which the graph crosses the vertical axis, which the vertex form conceals completely, and it shows $a > 0$, so both arms point up.

No new information entered anywhere in that expansion. One parabola, two costumes, and a different question answered for free in each.

Read the coefficients off the standard form: $a = \tfrac{1}{2}$ and $b = -8$. The axis is where symmetry puts it,

$x = -\frac{b}{2a} = -\frac{-8}{2 \cdot \tfrac{1}{2}} = \frac{8}{1} = 8,$

with both minus signs counted and the division by $2a = 1$ carried out rather than assumed. One evaluation gives the height there:

$g(8) = \tfrac{1}{2}(64) - 8(8) + 38 = 32 - 64 + 38 = 6,$

so the vertex is $(8, 6)$, agreeing with the vertex form we were told to ignore. Since $a > 0$ the graph opens upward, that height is the smallest output, and the range is $[6, \infty)$.

For the comparison with the parent, use the identity about the axis, which here reads $g(8 + t) = 6 + \tfrac{1}{2}t^2$. A displacement $t$ from the axis buys a rise of $\tfrac{1}{2}t^2$, where the parent buys $t^2$ from its own vertex: half as much at every displacement. So these arms climb less steeply and the parabola looks wider than $y = x^2$.

The standard form knew the vertex all along. It simply refused to display it.

Write $p(x) = x^2$ and read the rule as the parent with three constants attached.

$g(x) = \tfrac{1}{2}\,p(x - 8) + 6$

The inside change acts on the input before the parent runs, so it moves the graph right $8$, the direction opposite to the sign written inside. The $\tfrac{1}{2}$ and the $6$ act on the height the parent returns, scaling it and then raising it. In order: shift right $8$, scale every height by $\tfrac{1}{2}$, shift up $6$.

The landing rule from Transformations of Graphs sends a parent point $(u, v)$ to $(u + h,\ av + k)$, so

$(2, 4) \ \longmapsto \ \left(2 + 8,\ \tfrac{1}{2}(4) + 6\right) = (10, 8).$

Confirm against the rule itself, which is the only check that cannot be fooled by a misread move:

$g(10) = \tfrac{1}{2}(10 - 8)^2 + 6 = \tfrac{1}{2}(4) + 6 = 8.$

The vertex is the same statement about the parent's own vertex: $(0, 0)$ lands at $(8, 6)$, which is exactly what the vertex form announced in part A.

Two of the records return the same height, $-16$, at two different inputs. A vertical mirror pairs points of equal height at equal displacements either side of it, so $-7$ and $-1$ are partners and the mirror stands midway between them:

$x = \frac{-7 + (-1)}{2} = -4.$

The third record sits at exactly that input, so it is the height on the axis, and the point of a parabola on its axis is the vertex: $(-4, 11)$.

The direction now follows from a comparison, with no coefficient computed. The vertex is either the lowest or the highest point of the whole graph. Here the height on the axis is $11$, while $3$ units either side the graph is down at $-16$, so the vertex cannot be the lowest point. It is the highest, the parabola opens downward, and $a < 0$.

With the axis at $m = -4$ the identity reads $f(-4 + t) = 11 + at^2$. The record at $x = -1$ is the case $t = 3$:

$-16 = 11 + 9a \ \Longrightarrow \ 9a = -27 \ \Longrightarrow \ a = -3.$

The record at $x = -7$ is the case $t = -3$ and produces the same equation, since only $t^2$ appears. That is why one of the pair was enough, and the negative value agrees with the downward opening argued in part A.

Vertex form is now complete, and expanding it produces the standard form:

$f(x) = -3(x + 4)^2 + 11 = -3(x^2 + 8x + 16) + 11$

$= -3x^2 - 24x - 48 + 11 = -3x^2 - 24x - 37.$

Check every record against the expanded form, since that is the form the checking is cheapest in: $f(-7) = -147 + 168 - 37 = -16$, $f(-1) = -3 + 24 - 37 = -16$, and $f(-4) = -48 + 96 - 37 = 11$. All three hold, so the function reproduces each fact it was built from.

Take the two surviving records and ask what every quadratic satisfying them must share. If $f(-7) = f(-1)$, those inputs are a pair of equal heights, so they are mirror partners and the axis bisects them at $x = -4$, whatever the coefficients turn out to be. That much is forced.

It runs in the other direction too, which is worth saying out loud: a quadratic whose axis is $x = -4$ satisfies $f(-4 + t) = f(-4) + at^2$, and the inputs $-7$ and $-1$ are the cases $t = -3$ and $t = 3$, so their heights agree automatically. Axis at $-4$ and equal heights at those two inputs are the same condition, not one merely implying the other.

Nothing beyond the axis is forced. Fix the axis and the shared height, and let the leading coefficient be whatever it likes: every member of the family

$f_a(x) = a(x + 4)^2 - 16 - 9a, \qquad a \ne 0,$

has its vertex on $x = -4$ and returns $-16$ at $t = \pm 3$, that is at $-7$ and at $-1$. Two members make the point:

$a = -3: \ -3(x + 4)^2 + 11, \qquad a = 1: \ (x + 4)^2 - 25.$

The first opens downward with greatest value $11$; the second opens upward with least value $-25$. Both fit the surviving records exactly, and they disagree about the direction, the vertex height and the range. So the lost record was carrying all of that, and a pair of equal values on its own carries only the mirror line.

Expand the brochure record and set it beside the installer's, term by term.

$-\tfrac{1}{20}(x - 60)^2 + 180 = -\tfrac{1}{20}(x^2 - 120x + 3600) + 180$

$= -\tfrac{1}{20}x^2 + 6x - 180 + 180 = -\tfrac{1}{20}x^2 + 6x$

The two records are the same function, so they describe the same jet; the fountain does not care which sheet you read.

What differs is what each hands over for nothing. The brochure form is $a(x - h)^2 + k$ with $h = 60$ and $k = 180$, so it announces the top of the arc and the distance at which it happens. The installer's form is $ax^2 + bx + c$ with $c = 0$, so it announces the height at the nozzle, where $x = 0$: the water leaves at height $0$, the reference the whole model is measured from. The brochure form conceals that, and the installer's form conceals the top.

The brochure form does the first question with no work. It is $a(x - h)^2 + k$ with $a = -\tfrac{1}{20} < 0$, so the parabola opens downward and the vertex is the highest point of the whole arc:

$h(60) = 180.$

The jet reaches $180$ centimetres above the nozzle, at a horizontal distance of $60$ centimetres.

The second question looks like an equation to solve, and is not one. The installer's form gives the height at the nozzle at sight, $h(0) = c = 0$. The axis stands at $x = 60$, so the input $0$ sits $60$ to the left of it and the input $120$ sits $60$ to the right: equal displacements from the axis, so equal heights.

$h(60 + t) = 180 - \tfrac{1}{20}t^2, \qquad h(0) = h(120) = 180 - \tfrac{1}{20}(3600) = 0$

So the water is back at nozzle height $120$ centimetres from the nozzle. The two facts that made it free were the height at the nozzle, read off one record, and the position of the axis, read off the other. Neither was worked for.

Test the note by asking the model itself. The distance $30$ is $30$ to the left of the axis, and either record answers:

$h(30) = -\tfrac{1}{20}(900) + 6(30) = -45 + 180 = 135.$

The height there is $135$ centimetres, where the note predicts $90$, so the note is out by $45$ centimetres.

The assumption behind it is that height is gained at a steady rate, so that covering half the horizontal distance to the top buys half the height. That is how a straight ramp behaves. The identity about the axis says what this graph does instead:

$h(60 + t) = 180 - \tfrac{1}{20}t^2.$

The drop below the top is built from $t^2$, so halving the distance from the axis quarters the drop. At $t = 60$, the nozzle, the drop is the full $180$; at $t = 30$ it is $45$, a quarter of that, leaving $135$, which is three quarters of the top height rather than half of it. The water climbs steeply out of the nozzle and flattens as it nears the top, and that changing steepness is exactly the bend the note ignored.

Expand, keeping every constant as a letter so that the conclusion is not tied to one parabola.

$a(x - h)^2 + k = a(x^2 - 2hx + h^2) + k$

$= ax^2 + (-2ah)x + (ah^2 + k)$

So in standard form $b = -2ah$ and $c = ah^2 + k$, while the leading coefficient is the same $a$ it started as. Feed that $b$ into the axis formula, dividing by $2a$, which is legal precisely because $a \ne 0$:

$-\frac{b}{2a} = -\frac{-2ah}{2a} = h.$

The height there costs one evaluation of the vertex form, in which the squared term dies:

$f(h) = a(h - h)^2 + k = k.$

Nothing was assumed about $a$, $h$ or $k$ beyond $a \ne 0$, so this holds for every quadratic written in vertex form at once, rather than for the ones anybody happens to test. The two costumes therefore never disagree: whichever you are handed, the axis is the same line and the vertex is the same point, so a computation made in one form can always be checked in the other.

That settles the traffic in one direction, and the other direction is already paid for. For any $f(x) = ax^2 + bx + c$ with $a \ne 0$, write $m = -\frac{b}{2a}$; the identity about the axis proved in the lesson says $f(m + t) = f(m) + at^2$ for every displacement $t$. Every input is the axis plus some displacement, so take $t = x - m$:

$f(x) = a(x - m)^2 + f(m).$

That is a vertex form, assembled out of the coefficients alone. So the vertex form is not a costume that some quadratics happen to own and others do not: every quadratic has one, with $h = -\frac{b}{2a}$ and $k$ the height there.

Find the axis and the height there, then bound every output at once. With $a = 5$ and $b = -30$,

$x = -\frac{-30}{2 \cdot 5} = 3, \qquad f(3) = 45 - 90 + 52 = 7.$

The identity about the axis now describes the entire graph in one line:

$f(3 + t) = 7 + 5t^2.$

Every real $t$ makes $5t^2 \ge 0$, so $f(x) \ge 7$ for every real $x$. In particular $f$ never returns $0$.

Now suppose, for the sake of argument, that real numbers $r_1$ and $r_2$ existed with

$f(x) = 5(x - r_1)(x - r_2) \quad \text{for every } x,$

the leading coefficient being that same $5$, since expanding such a product starts $5x^2$. Substituting the number $r_1$ makes the first bracket $0$, and a product with a factor of $0$ is $0$:

$f(r_1) = 5(r_1 - r_1)(r_1 - r_2) = 0.$

That contradicts $f(x) \ge 7$. So no such pair of real numbers exists: this quadratic has no real factored form and no real zeros. Its graph opens upward with its lowest point $7$ above the horizontal axis, so it never reaches that axis to cross it.

Suppose the two rules are the same function, so they agree at every input:

$a(x - h)^2 + k = A(x - H)^2 + K \quad \text{for every } x.$

Part A puts the axis of the left rule at $x = h$ and the axis of the right rule at $x = H$. The lesson showed a quadratic has exactly one mirror line, so one graph cannot carry two different ones, and

$h = H.$

Now evaluate both rules at that common input. The squared term dies on each side, leaving

$k = K.$

Finally evaluate both at the input $h + 1$, one unit out from the axis. The left rule gives $a + k$ and the right gives $A + K$, and the constants just shown equal cancel:

$a + k = A + K \ \Longrightarrow \ a = A.$

All three constants match, so the triple was never free to vary: every quadratic has a vertex form, by part A, and has exactly one. That is what lets us speak of 'the vertex' and 'the axis' of a parabola at all, rather than of a vertex somebody happened to choose. Notice how different this is from part B: there the costume was unavailable, here it is available in exactly one way.

Take the crossings of the horizontal axis first. Substituting $x = -5$ makes the bracket $x + 5$ equal to $0$, and a product with a factor of $0$ is $0$:

$f(-5) = 2(0)(-6) = 0, \qquad f(1) = 2(6)(0) = 0.$

So the graph meets the horizontal axis at $x = -5$ and at $x = 1$, and both numbers were legible in the brackets before any arithmetic was done. There are no further crossings either: the identity about the axis makes the height at displacement $t$ equal to the height on the axis plus $2t^2$, which grows strictly as $|t|$ grows, so no height at all is taken at more than two inputs, and for the height $0$ those two inputs are already spoken for.

The crossing of the vertical axis is the one number this form does not display. It costs a single evaluation, at $x = 0$:

$f(0) = 2(0 + 5)(0 - 1) = 2(5)(-1) = -10.$

So the graph passes through $(-5, 0)$, $(1, 0)$ and $(0, -10)$.

The two crossings are a pair of inputs returning the same height, namely $0$, so they are mirror partners and the axis bisects them:

$x = \frac{-5 + 1}{2} = -2.$

One evaluation gives the height there, and the factored form is the cheapest place to do it:

$f(-2) = 2(-2 + 5)(-2 - 1) = 2(3)(-3) = -18,$

so the vertex is $(-2, -18)$. Now expand and check that route against the coefficients:

$f(x) = 2(x^2 + 4x - 5) = 2x^2 + 8x - 10,$

$x = -\frac{b}{2a} = -\frac{8}{4} = -2.$

The two routes agree, as they are bound to. Since $a = 2 > 0$ the parabola opens upward, so the vertex height is the smallest output the function takes and the range is $[-18, \infty)$. The expansion also hands back the crossing of the vertical axis a second time, as $c = -10$, matching the evaluation in part A.

Let $m$ be the axis. Every input can be written $m + t$ for exactly one displacement $t$, and the identity about the axis says

$f(m + t) = f(m) + at^2,$

so a height depends on its input only through $t^2$. Two distinct inputs $m + t_1$ and $m + t_2$ therefore have equal heights exactly when $at_1^2 = at_2^2$, and since $a \ne 0$ that means $t_1^2 = t_2^2$, which for distinct inputs forces $t_2 = -t_1$. The two crossings share the height $0$, so they are $m - t$ and $m + t$ for a single displacement $t$, and their midpoint is

$\frac{(m - t) + (m + t)}{2} = m.$

The axis is the midpoint of the crossings, and that argument never touched an expanded coefficient or the value of $a$: it used only that $a \ne 0$ and that a height is built from the square of the displacement.

Now replace the $2$ by $-2$, giving $g(x) = -2(x + 5)(x - 1)$. The brackets are untouched, so $g(-5) = 0$ and $g(1) = 0$ still: neither crossing moves, and neither does their midpoint, so the axis is still $x = -2$. Everything the sign of $a$ governs does change. The height on the axis becomes

$g(-2) = -2(3)(-3) = 18,$

the parabola opens downward, so that height is now the largest output rather than the smallest and the range becomes $(-\infty, 18]$; the crossing of the vertical axis flips as well, to $g(0) = -2(5)(-1) = 10$; and expanding now reverses every coefficient, giving the standard form $-2x^2 - 8x + 10$. Those are all one change wearing four hats, since $g(x) = -f(x)$ at every input: each output simply reverses sign, which leaves the two inputs of height $0$ exactly where they were.

So the sign of the leading coefficient sets the direction and which end of the range is closed, and flipping it, as this replacement does, reverses every coefficient of the standard form. The size of the leading coefficient is a separate matter: it sets the steepness, which does not change here at all, because $-2$ and $2$ are the same size. Where the mirror stands is settled by the crossings alone, and neither the sign nor the size can move it.