Look at the piece (x − h)². Whatever x you feed in, you are squaring a real number, so
Now multiply by a and add k. If a > 0, multiplying by a positive number keeps the direction of the inequality:
If a < 0, multiplying by a negative number reverses the inequality:
Either way the extreme value is k and it happens at x = h. That single point (h, k) is the vertex, and the sign of a alone decides whether it is the bottom of a valley or the top of a hill.
Take any distance d and step that far to the right of h, then that far to the left. Compare the two heights:
Identical — because (−d)² and (d)² are the same number. Squaring destroys the sign, so the graph cannot tell left from right. Every pair of points the same distance from x = h sits at the same height, which is exactly what a mirror line means. Switch the figure above to Symmetry to see the matched pairs drawn in.
This is also the fastest way to get free points when graphing: find one point, reflect it across x = h, and you have a second one without evaluating anything.
Vertex form and standard form are the same function wearing different clothes. Expand the vertex form and see which coefficient lands where:
Compare that with ax² + bx + c. The x² coefficients already match, so line up the other two:
So the “axis formula” is not a separate idea — it is what h looks like after the parentheses are multiplied out. Notice that both results divide by a, which is the first sign that a ≠ 0 is not optional.
The expansion above rests on (x − h)² = x² − 2hx + h². Here is why, with no algebra at all. Start with a square of side x, area x², and trim a strip of width h off the right edge and another off the bottom:
The two gold strips each have area h · x, so removing both takes away 2hx. But look at the darker corner: it belongs to both strips, so trimming both removed it twice. Put it back once. What is left is the blue square of side x − h:
That stray +h² is the corner you added back. It is also, not coincidentally, the number you add when you complete the square — case 3 in the stepper uses it.
Nearly everyone expects y = (x − 3)² to shift left. It shifts right. The reason: the graph does whatever it does when the inside of the square is a particular number, and subtracting 3 means x must be 3 units larger to produce that same inside value.
| inside is… | −2 | −1 | 0 | 1 | 2 |
|---|---|---|---|---|---|
| y = x² needs x = | −2 | −1 | 0 | 1 | 2 |
| y = (x − 3)² needs x = | 1 | 2 | 3 | 4 | 5 |
| height y | 4 | 1 | 0 | 1 | 4 |
Same heights, in the same order, every x three units later — the graph moved 3 to the right. And k holds no such surprise: it is added after the squaring, directly to y, so +k lifts the graph and −k drops it, exactly as written.
Start at the vertex and step 1 unit sideways. The square contributes 1, so the height changes by exactly a. Step 2 units and the square contributes 4, so the height changes by 4a. Every rise is scaled by a:
| from the vertex, step… | a = 1 | a = 3 | a = ½ | a = −2 |
|---|---|---|---|---|
| 1 unit sideways, rise | 1 | 3 | 0.5 | −2 |
| 2 units sideways, rise | 4 | 12 | 2 | −8 |
| 3 units sideways, rise | 9 | 27 | 4.5 | −18 |
|a| > 1 makes every rise bigger, so the curve climbs faster and looks narrower. 0 < |a| < 1 makes every rise smaller, so it looks wider. A negative a turns every rise into a fall, flipping the whole graph across its own horizontal line through the vertex. The 1, 4, 9 pattern in the first column is worth memorizing on its own — it is the fastest hand-plotting method there is: from the vertex, go over 1 and up a, over 2 and up 4a, over 3 and up 9a, then mirror.
Domain: all real numbers. A quadratic is a polynomial — there is no variable in a denominator and none under a radical — so there is nothing to exclude and no value of x is illegal. Write it as −∞ < x < ∞.
Range: decided entirely by a and k. From the inequality proved at the top, the range is y ≥ k when a > 0 and y ≤ k when a < 0. It never depends on h, because sliding a graph sideways changes no heights.
The one excluded value is a ≠ 0. Set a = 0 and the squared term vanishes: y = 0(x − h)² + k collapses to the horizontal line y = k. There is no vertex, no axis of symmetry in the usual sense, and no direction of opening — and the formula h = −b/2a would divide by zero, which is undefined rather than infinite. Drag a to 0 on the slider and watch every reading above go blank. Be alert when a holds a parameter: (m − 4)x² + 6x + 1 stops being a parabola at exactly m = 4.
Not every parabola has x-intercepts. Solving a(x − h)² + k = 0 gives (x − h)² = −k/a, and a square cannot equal a negative number. So real x-intercepts exist only when −k/a ≥ 0 — that is, when k = 0, or when a and k have opposite signs. In pictures: a parabola opening up must have its vertex at or below the x-axis, and one opening down must have its vertex at or above it. Case 4 in the stepper is an example with none.
On extraneous solutions. Taking the square root of both sides is safe here provided you keep the ±: (x − h)² = m is equivalent to x − h = ±√m, so the step invents nothing false — dropping the ± loses a true root rather than adding a fake one. Extraneous roots come from elsewhere. If your quadratic appeared after clearing a denominator, any root that made that denominator zero was never legal and must be discarded no matter how clean it looks. If it appeared after squaring both sides of a radical equation, every root must be substituted back into the original equation, since squaring can turn a false statement true. And in an applied problem, a root that means a negative length or a negative time is rejected on meaning, not on algebra. The graph shows every root the quadratic has; only the original problem decides which ones you are allowed to keep.