Quadratic Functions & Parabolas

A quadratic is a shape

Throw a ball. Arc a fountain of water. Track a diver from the board to the splash. Every one of those paths is the same shape — a parabola — and every parabola is the picture of a quadratic. M2 taught you to handle $x^2 + 5x + 6$ as an expression to factor; M3 taught you to solve it when it equals zero. This module gives that expression a body: turn it into a function $f(x) = x^2 + 5x + 6$ and plot every input against its output, and you get a curve you can read.

The simplest quadratic, $y = x^2$, is the parent parabola: a symmetric U resting its bottom at the origin. Every other parabola is that same U, moved and restretched. The leading number $a$ does two jobs at once: its sign decides which way the U opens — positive is a smile (a lowest point), negative is a frown (a highest point) — and its size decides how narrow. A big $|a|$ pinches the U tight; a small one lets it flare wide open. That single coefficient is why one quadratic launches steeply and another drifts.

Three forms, three pairs of glasses

Here is the idea the whole module turns on. A parabola can be written three ways, and they are all the same curve — the same points, the same picture. What changes is which feature each one hands you for free, as a number you can read straight off without doing any work. They are three pairs of glasses on one shape.

• Vertex form $a(x - h)^2 + k$ shows the turning point. The vertex — the peak or the valley — sits at $(h, k)$, right there in the constants.
• Factored form $a(x - r_1)(x - r_2)$ shows the x-intercepts. The graph crosses the axis at $r_1$ and $r_2$.
• Standard form $ax^2 + bx + c$ shows the y-intercept. Set $x = 0$ and every $x$-term vanishes, so the curve crosses the y-axis at $c$.

Say a problem hands you standard form and asks for the vertex. You convert to the form that shows it — by completing the square, the M2 move, now with an $a$ to carry:

Factored form is the flip side. If $f(x) = (x - 3)(x + 1)$, the graph crosses where each factor is zero: at $x = 3$ and $x = -1$. Notice both signs flipped — the bracket $(x - 3)$ hits zero at $+3$, not $-3$. That sign flip is the same one M3 warned about when reading roots off factored form, and it is worth points every time.

Feel all three at once by dragging. In Parabola Studio, move the vertex and watch every form rewrite itself, then tap a form to light up the feature it owns.

Predict before you click: load just touches, then tap “the x-intercepts.” How many crossing dots light up? Now drag the vertex one step down — do the two crossings appear, and what does $b^2 - 4ac$ read now? Then tap “the vertex” and drag sideways: only the vertex form’s numbers should follow your hand cleanly, because that’s the form built to track it.

Where the parabola meets the axis

M3 promised you would see the discriminant here, and this is it. The x-intercepts are exactly the solutions of $f(x) = 0$ — so the number of times a parabola crosses the x-axis is the number of real solutions its equation has, the verdict $b^2 - 4ac$ already knew:

• $b^2 - 4ac > 0$: two real solutions, so the parabola crosses the axis twice.
• $b^2 - 4ac = 0$: one solution, so the parabola just touches the axis — its vertex sits exactly on it.
• $b^2 - 4ac < 0$: no real solution, so the parabola misses the axis entirely, floating above it or below it.

Lift $y = x^2$ upward by adding to it — $x^2$, then $x^2 + 1$, then $x^2 + 4$ — and you watch its two crossings slide together, kiss, and lift off the axis. The three cases M3 counted are three heights of the same curve. Watch it happen at the board:

Every parabola is also mirror-symmetric about a vertical line through its vertex: the axis of symmetry, $x = -\tfrac{b}{2a}$. That formula is the fastest route to a vertex you’ll ever need — find the axis, then feed that $x$ back into the function for the $y$. It’s also why the two x-intercepts, when they exist, always sit an equal step to each side of the vertex: the $\pm$ from M3’s formula spreads them evenly around $-\tfrac{b}{2a}$, the same balance point the twins never left in the Discriminant Lab.

The other direction — building a parabola from instructions — is where vertex form really sings. Read $a(x - h)^2 + k$ as a recipe applied to the parent $y = x^2$: shift right $h$, shift up $k$, stretch by $a$ (and flip if $a$ is negative). Shift & Stretch lets you slide those three dials and watch the parent move.

Predict before you slide: set $h$ to $+3$. The bracket now reads $x - 3$ — a minus — so which way did the curve go? (Right — the trap from earlier, now under your finger.) Then push $a$ past zero into the negatives and watch the smile become a frown; slide it to $\tfrac12$ and watch the U flare wide.

The one thing to remember

A parabola is one shape you can write three ways, and you pick the form that already shows what you’re asked for: vertex form for the peak or valley, factored form for the crossings, standard form for the y-start. When you only need how many times it crosses, don’t graph — the discriminant $b^2 - 4ac$ tells you, just as it did in M3. And the axis of symmetry $x = -\tfrac{b}{2a}$ is always the shortcut home to the vertex.

Pick the form that shows what you need

Opening & width: $a > 0$ opens up (a lowest point, a minimum); $a < 0$ opens down (a highest point, a maximum). Bigger $|a|$ is narrower, smaller is wider.

How many x-intercepts — the discriminant, no graphing: $\;b^2 - 4ac > 0$ → two (crosses). $\;= 0$ → one (touches at the vertex). $\;< 0$ → none (misses the axis).

Complete the square (a ≠ 1) to reach vertex form: factor $a$ out of the $x$-terms, add and subtract $\left(\tfrac{b/a}{2}\right)^2$ inside, and carry the leftover out multiplied by $a$.