Nonlinear Systems & Transformations

Two graphs, one plane

Put a parabola and a line on the same grid and one question writes itself: do they meet? A solution of a system is a point that makes both equations true at once — which on a graph means a point sitting on both curves (A6’s idea, now with a bend in it). Elimination is useless here — you can’t cancel an $x^2$ against an $x$ — but nothing stops you from noticing that at a shared point, the two recipes for $y$ must produce the same number:

Check either point against the parabola and it agrees — that’s what “on both graphs” means. One substitution, and every line-meets-curve question on the SAT opens the same way.

One number counts the crossings

Here is where M3’s discriminant earns a second life. The equation $x^2 - 6x + 8 = 0$ had $D = b^2 - 4ac = 36 - 32 = 4$ — positive, two solutions, two crossings. But nothing forced the line to cross twice. Slide it down and the two shared points slide toward each other; at exactly one height they merge into a single graze; lower still, no contact at all. Algebraically that is just $D$ passing through zero: $D > 0$ cross, $D = 0$ touch, $D < 0$ miss. The discriminant stops counting roots and starts counting intersections.

It opens on the system above, crossings at $(2, 1)$ and $(4, 5)$. The two green dots have one height where they meet — predict its $k$ before you drag, then pull the k dot down one step at a time and watch $D$ shrink: $4$, then $0$ — the kiss, at $(3, 2)$ — then negative, and the dots are gone. Try opens down too: with the parabola hanging, it’s dragging the line up that makes it miss.

The k the SAT asks for

“For what value of $k$ does $y = 2x + k$ intersect $y = x^2 - 4x + 5$ at exactly one point?” — the exam’s favorite dress for everything above. Run the same substitution with $k$ left as a letter: $x^2 - 6x + (5 - k) = 0$. Exactly one intersection means exactly one solution, and that is a condition on the discriminant: $(-6)^2 - 4(1)(5 - k) = 0$, so $36 - 20 + 4k = 0$, so $16 + 4k = 0$ and $k = -4$ — precisely the height where your drag found the kiss.

What about two curves? Set $y = x^2 - 2x + 3$ against $y = x^2 + 4x - 9$: the recipes equal, $x^2 - 2x + 3 = x^2 + 4x - 9$ — and both sides carry the same $x^2$, so subtracting wipes the squares out entirely: $-6x + 12 = 0$, $x = 2$, and either curve gives the height $y = 3$. The scary-looking system was secretly linear.

Any graph, four moves

The second half of this unit is a grammar — four small edits to a function’s formula, each with a fixed geometric meaning. You’ve already seen it once in a special costume: vertex form sliding a parabola around (M4’s story). The SAT’s twist is that the grammar works on any graph, including ones you’re handed with no formula at all. Suppose all you know is that $(2, 5)$ sits on the graph of $f$, and $g(x) = f(x - 1) + 3$. Which point must sit on $g$?

The four moves, then: $f(x) + k$ slides the graph up $k$; $a \cdot f(x)$ stretches every height by $a$; $-f(x)$ flips it over the x-axis — all three act on the output, so they do exactly what they say. Only $f(x - h)$ is sneaky: it acts on the input, and input changes run backwards — minus inside means right.

It opens with the copy lying on its dashed ghost and the readout at $g(x) = f(x)$. Before you drag: pull the dot $3$ to the right — what will the parentheses say, plus or minus? Check, then drag $2$ up and watch the $+2$ land outside. Now reset and tap − flip: the gray marked point of $f$ is $(3, -2)$ — predict where the green dot must land before you look. Flip plus ×2 sends it to $(3, 4)$; the mapping line does the arithmetic with you.

The one thing to remember

A shared point makes both equations true, so set the two recipes for $y$ equal — one substitution turns any line–curve system into a single quadratic, whose discriminant counts the crossings before you solve: miss, touch, cross. And any graph, formula or not, moves by four small edits: outside the parentheses is honest to $y$, inside is backwards to $x$.

Line meets curve

Solve the system: set the two expressions for $y$ equal → move everything to one side → solve the quadratic → get each $y$ from the line. Answer with the point(s) — the SAT often asks for one coordinate.

“For what $k$ … exactly one point”: keep $k$ symbolic, set the one-side discriminant to $0$, solve for $k$. Exactly the same condition as “exactly one solution”.

Two parabolas, same $x^2$ coefficient: setting them equal cancels the squares — the system is linear. Solve the one-step equation; the height comes from either curve.

The four moves

Point rule: $(x, y)$ on $f$ $\Rightarrow$ $(x + h,\; a\,y + k)$ on $g$. Reverse it to walk a point of $g$ back to $f$: $(x - h,\; (y - k)/a)$.