From y = ax² to vertex form y = a(x−h)²+k: which way it opens, how wide, and where the bowl sits.
Toss a ball straight out and watch its path arc up and back down — that curve is a parabola, the shape of every quadratic graph. Starting from the cleanest case y = ax², this lesson learns to read the bowl. The sign of a decides whether it opens up or down; the size of a decides how narrow or wide it is; and two simple moves — adding a number, or replacing x by x − h — slide the whole bowl around the plane without bending it. Put the two moves together and you reach vertex form, where the turning point is written right into the equation. By the end you will be able to sketch any quadratic, and read any quadratic's picture, almost at a glance.
Every quadratic function y = ax² + bx + c graphs as a parabola, and the simplest of all is y = ax² — no b, no c, just the squared term. Its turning point sits right at the origin. The first thing to read off any parabola is which way it opens, and that is decided by one thing alone: the sign of a.
When a > 0, squaring makes the output grow as you move away from the centre in either direction, so the bowl opens upward — it could hold water, and its vertex is the lowest point. When a < 0, every output is flipped below the axis, so the bowl opens downward — like a fountain's arch — and its vertex is the highest point. You meet both shapes every day: a satellite dish curves up to gather signals at its lowest point; a thrown ball traces a path that curves down to a highest point and falls back.
For y = ax² (and for every quadratic): if a > 0 the parabola opens upward and the vertex is a minimum; if a < 0 it opens downward and the vertex is a maximum. The case a = 0 is forbidden — with no squared term you would have a straight line, not a parabola.
The sign of a tells you up or down; the size of a — that is, |a| — tells you the shape. The bigger |a| is, the faster the outputs climb away from the vertex, so the bowl is narrower and steeper. The smaller |a| is, the more gently it rises, so the bowl is wider and flatter. The parent y = x², with |a| = 1, is the dividing line.
Check it with one input. At x = 2: the wide bowl y = ½x² reaches only 2; the parent y = x² reaches 4; the narrow bowl y = 2x² already reaches 8. Same input, taller and taller climbs — that is exactly what "narrower" looks like. Through it all the vertex never budges: it stays parked at the origin. Changing a reshapes the bowl; it does not move it.
Slide a from −3 to 3. Watch the bowl flip when a passes 0, and narrow as |a| grows. The slate ghost is the parent y = x² for comparison. (a = 0 is skipped — it would be a straight line.)
Larger |a| makes a parabola narrower, not "taller" — the whole curve squeezes in toward its axis. And for y = ax² the vertex stays at the origin no matter how you change a. To move the bowl you need the shifts coming next.
Now we move the bowl. Take the parent y = x² and add a number k to every output: y = x² + k. Every point rises by the same k, so the whole curve lifts straight up (when k > 0) or drops straight down (when k < 0), keeping its exact shape. The vertex, once at (0, 0), lands at (0, k).
Worked. y = x² + 2 is the parent lifted up 2; the vertex moves to (0, 2). And y = x² − 3 is the parent dropped down 3; the vertex moves to (0, −3). The bowl is identical in both — only its height has changed.
The graph of y = ax² + k is the bowl of y = ax² slid vertically by k: up for k > 0, down for k < 0. Its vertex is (0, k) and its axis is still the y-axis. The shape — opening and width — is untouched.
Vertical shifts were easy because we changed the output. Horizontal shifts work on the input, and they have a famous twist. Replace x by x − h in the parent: y = (x − h)². The lowest output, 0, now happens not at x = 0 but where the inside is zero — that is, where x − h = 0, i.e. at x = h. So the whole bowl slides h to the right (for h > 0), and its vertex lands at (h, 0).
Worked. y = (x − 2)² slides the parent right 2; vertex (2, 0). Going the other way, y = (x + 3)² is really y = (x − (−3))², so h = −3 and the bowl slides left 3; vertex (−3, 0). The sign feels backward — "x minus" moves right — so always read it as "the vertex sits where x = h."
y = (x − h)² moves the bowl RIGHT by h, and y = (x + h)² moves it LEFT. It is easy to slip and shift the wrong way. The cure: don't memorize "minus means right" — instead set the inside to zero. The vertex is wherever x − h = 0, namely x = h.
With a = 1 fixed, dial h (left/right) and k (up/down). The slate ghost is the parent y = x²; the slate arrow runs from O to the new vertex (h, k). Notice the shape never changes — only the position.
Why settle for one shift when you can do both? Start from y = ax², replace x by x − h to slide it sideways, and add k to slide it up — the bowl lands with its vertex at (h, k) and its axis at x = h. The single equation that records all of this is
y = a(x − h)² + k.
Because the turning point (h, k) can be read straight off the equation, this is called vertex form. The leading a still does its two jobs — its sign opens the bowl up or down, its size makes it narrow or wide — while h and k place the vertex anywhere on the plane.
Worked. Read y = 2(x − 1)² + 3. Here a = 2 (positive, so it opens up, and |a| = 2 > 1 so it is narrow); h = 1 and k = 3, so the vertex is (1, 3) and the axis is x = 1. That one glance gives the entire sketch.
Dial a (sign = up/down, size = narrow/wide), h (axis position), and k (vertex height). The green dot is the vertex (h, k); the dashed slate line is the axis x = h. Everything in the readout is computed from a, h, k. (a = 0 is skipped.)
In y = a(x − h)² + k the turning point is (h, k) and the axis of symmetry is x = h. The sign of a sets the opening (up if a > 0, down if a < 0); the size |a| sets the width (narrow if |a| > 1, wide if |a| < 1). When you see "x + 3", rewrite it as "x − (−3)" to read off h = −3.
A quadratic graphs as a parabola, and every parabola is a reshaped, shifted copy of the parent y = x²:
Next (23.3) we take a quadratic in everyday general form y = ax² + bx + c and recover that same vertex and axis with the formula x = −b2a and by completing the square.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops the graph of a quadratic as a transformation of the parent function, leading to vertex form. It aligns with F-BF.B.3 (identify the effect of replacing f(x) by a·f(x), f(x)+k, and f(x−h) — vertical stretch/reflection and translation), F-IF.C.7a (graph quadratic functions, showing the vertex), and A-SSE.B.3b (vertex form reveals the maximum or minimum). It builds on Stage 20's coordinate plane and function ideas and feeds directly into 23.3, where the same vertex and axis are extracted from general form.
Common misconceptions to watch for. Many students think y = (x − h)² shifts left because of the minus sign — counter this by asking "what x makes the inside zero?" so they read the vertex as x = h. Others believe a larger a makes the parabola "taller" rather than narrower; emphasize that the vertex does not move when only a changes. Watch too for sign slips on k (subtracting instead of adding), and for the belief that a can be 0. Encourage learners to sketch from vertex form by plotting the vertex first, then one point on each side using symmetry.