Vertex form tells you everything. Standard form hides it one step away.
Find the vertex
What is the vertex of y = 2(x − 3)² + 5?
In y = a(x − h)² + k the vertex is (h, k). Here the bracket reads (x − 3), so h = 3, and k = 5, giving (3, 5). A is the standard trap — the minus in the formula means the sign flips, so (x − 3) gives +3, not −3. A bracket reading (x + 4) would give h = −4.
| Part | Controls | Detail |
|---|---|---|
| a | opening and width | positive opens up, negative down; larger |a| is narrower |
| h | left–right position | the x of the vertex; the sign flips |
| k | up–down position | the y of the vertex |
The graph of every quadratic function is a parabola. The axis of symmetry is the vertical line x = h through the vertex. The parabola is a mirror image on either side of it, which is why two x-intercepts are always the same distance from it.
The x-shift is inside the bracket and its sign reverses; the y-shift is outside and behaves normally. That asymmetry is the source of most errors here.
y = x² − 14x + 51
standard form
= (x² − 14x + 49) + 51 − 49
add and subtract (142)²
= (x − 7)² + 2
vertex (7, 2)
Adding 49 and subtracting it changes nothing, but it lets the first three terms fold into a perfect square. The leftover 2 is the vertex's y-value.
A shortcut for the vertex without all this: h = −b2a, then substitute to get k. Here −−142 = 7, and y = 49 − 98 + 51 = 2 ✓
The maximum value is k. Where it happens is h. Questions ask for one or the other, and they are different numbers.
y = x² − 2x − 8
find both intercepts
x = 0 → y = −8
the y-intercept
y = 0 → (x − 4)(x + 2) = 0
factor
x = 4 and x = −2
two x-intercepts
The axis of symmetry sits midway between them: 4 + (−2)2 = 1, which matches −b2a = 22 = 1 ✓ Two independent routes to the same line.
h = −b2a: the 2a matters as soon as a is not 1. And a vertex plus one point fixes the whole parabola.
Find the maximum of y = −2x² + 12x − 7.
h = −122 × −2 = 3
a = −2, b = 12
k = −2(9) + 36 − 7 = 11
substitute x = 3
maximum 11, at x = 3
a < 0, so the vertex is the top
Using −ba gives −12−2 = 6, and then y = −72 + 72 − 7 = −7. That point is on the parabola, but it is not the top: the real maximum, 11, is bigger.
A parabola has vertex (2, −3) and passes through (4, 5). Find its equation.
y = a(x − 2)² − 3
the vertex fills in h and k
5 = a(4 − 2)² − 3 = 4a − 3
the point fills in x and y
a = 2, so y = 2(x − 2)² − 3
check: 2(2)² − 3 = 5 ✓
Where the vertex sits decides how many times the graph meets the x-axis.
The x-intercepts of y = x² + 2x + 1
y = (x + 1)²
a perfect square
vertex (−1, 0)
the vertex is ON the axis
one x-intercept, x = −1
it touches and turns back
k = 0 means the parabola just touches the x-axis once.
The x-intercepts of y = −(x − 1)² − 4
a = −1, vertex (1, −4)
opens down from below the axis
highest value −4
never reaches y = 0
no x-intercepts
check: −(x − 1)² = 4 has no real x
Opening down with a negative top, or up with a positive bottom: no crossings.
The y-intercept of y = 3(x − 1)² − 5
y = 3(0 − 1)² − 5
set x = 0
y = 3 − 5 = −2
not −5
In vertex form the constant is k, the vertex height. The y-intercept needs x = 0 substituted.
Expect the vertex or axis of symmetry from either form, a maximum or minimum value, a shifted graph's equation, the intercepts, and sometimes an equation built from a vertex and one point. The fastest route is h = −b2a and then substitute for k; in vertex form, read (h, k) with the sign inside the bracket flipped. The 10-second check: the axis must sit midway between the two x-intercepts, and substituting h must give k.
Find the maximum
What is the maximum value of y = −3(x + 2)² + 11?
a = −3 is negative, so the parabola opens downward and the vertex is a maximum. The vertex is (−2, 11), and the maximum value is the y-coordinate: 11. A is where the maximum occurs, not the value itself — a distinction worth reading the question for.
Hint: check the sign of a to know whether the vertex is a maximum or a minimum, then read off k — the value is always the y-coordinate.
