1
Vertex form is a rewriting, not a new function
Expanding a(x - h)² + k returns ax² + bx + c. The two expressions agree at every x, so nothing has been gained or lost except readability — which is the only reason to do it.
Lesson · Quadratics and Complex Numbers · Parabolas
Vertex Form and the Axis of Symmetry
y = a(x - h)² + k is the same function as y = ax² + bx + c, written so that the vertex (h, k) and the axis x = h can be read without solving anything. Completing the square is the conversion, and in a function it is done by adding and subtracting rather than by adding to both sides.
y = a(x - h)² + k vertex (h, k), axis x = h
h = -b/(2a) k = f(h) = c - b²/(4a)
a > 0: k is the smallest value a < 0: k is the largest
2(x - 3)² - 5 is 2x² - 12x + 13 the same function
Vertex form, and where h and k come from
Core ideas
What this lesson establishes
One rewriting, and the two things it makes visible that standard form hides.
2
h is subtracted, so watch the sign
(x + 3)² is (x - (-3))², so h = -3 and the axis is x = -3. The form has a minus sign built in, and reading +3 off the bracket puts the vertex on the wrong side of the y-axis.
3
Keep the rule unchanged while y stays isolated
y = f(x) is an equation, so balance operations remain legal. But the goal here is another expression for the same output with y still by itself. Complete the square in f(x) = x² + 6x + 1 by adding 9 and subtracting 9 inside the expression; its value then stays unchanged at every x.
The material
Vertex form
What the form shows, why it shows it, and how to reach it.
Definition · Vertex
The vertex of a parabola is the point where it meets its axis of symmetry: the lowest point when the curve opens upward, the highest when it opens downward. A parabola has exactly one.
Definition · Vertex form
y = a(x - h)² + k, with a ≠ 0. The vertex is (h, k) and the axis of symmetry is the line x = h. The coefficient a is the same a as in standard form and does the same two jobs.
Theorem · Vertex form displays the extreme value
For f(x) = a(x - h)² + k: if a > 0 then f(x) ≥ k for every x, with equality only at x = h; if a < 0 then f(x) ≤ k for every x, with equality only at x = h.
x - h is a real number, so (x - h)² ≥ 0, and (x - h)² = 0 exactly when x = h. That is the whole content of the theorem; the rest is one multiplication.
If a > 0, multiplying (x - h)² ≥ 0 by a preserves the inequality: a(x - h)² ≥ 0. Adding k to both sides gives f(x) ≥ k, with equality exactly when (x - h)² = 0.
If a < 0, multiplying by a reverses it — the one rule of “Linear Inequalities” in Linear Equations and Inequalities — so a(x - h)² ≤ 0 and f(x) ≤ k, again with equality exactly at x = h.
□
That proof is worth more than the form. It says the vertex is the optimum without any appeal to a picture, and “Maximum and Minimum Problems” uses it on problems where there is no picture to appeal to.
Converting by completing the square
y = 2x² - 12x + 13
= 2(x² - 6x) + 13 factor 2 out of the x terms ONLY
= 2(x² - 6x + 9 - 9) + 13 half of -6 is -3, squared is 9
= 2((x - 3)² - 9) + 13
= 2(x - 3)² - 18 + 13 the -9 leaves the bracket as -18
= 2(x - 3)² - 5
vertex (3, -5) axis x = 3 a = 2 > 0, so -5 is the minimum
The fifth line is where this goes wrong. The -9 sits inside a bracket that is multiplied by 2, so bringing it out costs -18, not -9. Every conversion with a ≠ 1 has that step in it, and the check is instant: f(3) = 18 - 36 + 13 = -5, which is k.
Note what was not done. The constant 13 was left alone until the end, and the 9 was added and subtracted in the same line. Writing y + 9 = 2(x² - 6x + 9) + 13 looks like balancing but is not: the 9 inside the bracket carries the factor 2, so the right side rises by 18 while the left rises by 9, and the line defines y = 2x² - 12x + 22 — a different function. Adding and subtracting inside the expression is what removes the need to match two sides at all.
Where h and k come from
a(x - h)² + k = ax² - 2ahx + (ah² + k)
compare with ax² + bx + c
b = -2ah ⟹ h = -b/(2a)
c = ah² + k ⟹ k = c - ah² = c - b²/(4a)
and since k = f(h), computing f(h) directly is usually easier
So there are two routes. Completing the square shows the work and is the one procedure the course asks you to own; h = -b/(2a) followed by k = f(h) is quicker and involves fewer fractions. They agree, because the second was derived from the first.
Example · y = -2x² + 8x - 5
-2(x² - 4x) - 5 = -2((x - 2)² - 4) - 5 = -2(x - 2)² + 8 - 5 = -2(x - 2)² + 3. The vertex is (2, 3) and a = -2 < 0, so 3 is the largest value the function takes. Check: f(2) = -8 + 16 - 5 = 3.
Example · y = x² - 3x + 1
h = -b/(2a) = 3/2 and k = f(3/2) = 9/4 - 9/2 + 1 = -5/4, so y = (x - 3/2)² - 5/4. Halving an odd b produces halves and quarters; keep them. Multiplying through by 4 to clear them would define a different function, four times as tall.
Vertex form is also the transformation form of “Transformations of Graphs” in Lines, Functions and Graphs. y = a(x - h)² + k is the parabola y = x² stretched by a factor a, shifted h to the right and k up — which is why every parabola really is the same curve, seen at a different size and in a different place.
Interactive
Vertex form
a(x - h)^2 + k, derived by completing the square
Method
Converting ax² + bx + c to vertex form
Four steps. The third is the one that is skipped, and it is the one with a in it.
1
Factor a out of the x² and x terms only
2x² - 12x + 13 becomes 2(x² - 6x) + 13. The constant stays outside the bracket, untouched.
2
Halve the coefficient of x inside, and square it
Inside the bracket the coefficient is -6; half of it is -3 and the square is 9. That 9 is the number that completes the square, and it is the only number that does.
3
Add and subtract it inside, then take the subtraction out
2(x² - 6x + 9 - 9) is 2((x - 3)² - 9) is 2(x - 3)² - 18. The -9 is multiplied by a on the way out. Forgetting that multiplication is the standard error and it moves k by a times the square.
4
Read (h, k) and check with f(h)
The bracket (x - 3) gives h = 3; the constant left over gives k. Substituting h back into the original is a two-line check that catches everything above.
Vertex form
Score 0 / 0Choose the answer you can justify.
y = x² + 5x + 1 in vertex form
a = 1, so nothing has to be factored out and the arithmetic is purely in quarters.
y = x² + 5x + 1
half of 5 is 5/2, and (5/2)² = 25/4
= (x² + 5x + 25/4 - 25/4) + 1 add and subtract 25/4
= (x + 5/2)² - 25/4 + 1
= (x + 5/2)² - 21/4 since -25/4 + 4/4 = -21/4
vertex (-5/2, -21/4) axis x = -5/2
check h = -b/(2a) = -5/2 ✓
k = f(-5/2) = 25/4 - 25/2 + 1 = -21/4 ✓
The bracket reads (x + 5/2), so h = -5/2 and not 5/2. Vertex form subtracts h, and a plus sign inside the bracket always means a negative h.
Both halves of the check were computed from the original coefficients rather than from the vertex form, which is what makes the agreement worth something. a = 1 > 0, so -21/4 is the smallest value this function takes, and it takes it once.
For a faded conversion, rewrite y = 3x² + 12x + 5 in vertex form. The supplied first line is 3(x² + 4x) + 5; complete the square inside that bracket and account for the outside 3. Read the vertex and minimum, then check with h = -b/(2a) and f(h). Compare with y = 3(x + 2)² - 7 and vertex (-2, -7); a constant of -3 instead of -7 exposes the missing multiplication by a.
Common mistakes
1
Bringing the subtracted square out without multiplying by aIn 2(x² - 6x + 9 - 9) + 13 the -9 is inside a bracket carrying a factor of 2, so it emerges as -18. Writing 2(x - 3)² - 9 + 13 puts the vertex 9 units too high, and f(3) disagrees at once.
2
Reading h with the wrong sign(x + 3)² has its vertex at x = -3. The subtraction is part of the form, so a plus inside the bracket means a negative h — the same reversal that catches everyone in the transformations of Lines, Functions and Graphs.
3
Adjusting one side of a function definition onlyy + 9 = x² + 6x + 10 is still y = x² + 6x + 1, since subtracting 9 from both sides returns it, and 4y = 4x² + 24x + 4 divides back to it as well. The error is changing one side alone: y = x² + 6x + 10 adds the completing-square constant to the right only, and y = 4x² + 24x + 4 scales the right only, defining a function four times as tall. Inside a function definition the only legal move is one that leaves the value unchanged at every x, which is why completing the square there is add-and-subtract.
Check your understanding
✓
Finish when you can convert with a ≠ 1 and check it in two lines.
Take y = 3x² + 12x + 5, produce vertex form, and verify with h = -b/(2a) and k = f(h) computed from the original coefficients. If the two disagree, the factor a on the way out of the bracket is the first place to look.
Note: This is the same completing the square that derived the quadratic formula in “The Quadratic Formula”. One procedure, two payoffs: run it on ax² + bx + c = 0 and it gives the formula, run it on y = ax² + bx + c and it gives the vertex. “Maximum and Minimum Problems” spends that second payoff.