The highest or lowest point — and the two questions it answers.
A ball is thrown upward. Its height in metres after t seconds is
h = −5t² + 20t
How high does it get?
The peak is at t = −b2a = −20 ÷ (−10) = 2 seconds, and the height there is −5(4) + 40 = 20 m. D is the time, not the height — the vertex carries two coordinates and the question wants the second.
The sign of a decides whether the vertex is a maximum or a minimum.
opens downward ∩
the vertex is the highest point
opens upward ∪
the vertex is the lowest point
Here a = −5, so the curve opens downward and the vertex is a maximum. Good — a thrown ball does come back down.
The y-intercept is the value at x = 0, which is just c. y = −2x² + 5x + 4 opens downward (a = −2) and crosses the y-axis at 4.
This gives the time, not the height.
a
−5
b
20
c
0
t = −b2a
the axis of symmetry
t = −202(−5)
drop the values in
t = −20−10 = 2
two negatives make a plus
The ball peaks after 2 seconds.
−b2a is only half an answer. The height needs a substitution.
h = −5t² + 20t
the rule
h = −5(2)² + 20(2)
t = 2
h = −20 + 40
−5 · 4 = −20
h = 20
metres
“When” wants t = 2. “How high” wants h = 20. Giving 2 as the maximum height is the standard trap in exam questions like this.
The dashed line is the axis of symmetry. The curve is a mirror image either side of it. It is the vertical line through the vertex: here t = 2, and a parabola with vertex (3, −1) has axis x = 3.
Which explains a neat shortcut: the ball leaves the ground at t = 0 and lands at t = 4, and the peak sits exactly halfway between them.
y = a(x − h)² + k
vertex at (h, k)
h = −5t² + 20t
our equation
= −5(t² − 4t)
factor out the −5
= −5(t − 2)² + 20
complete the square
The 2 and the 20 are sitting right there — same answer, no formula needed.
(t − 2)² means the vertex is at t = +2. The bracket shows the opposite sign.
The vertex’s x does not depend on k, so find it first.
x = −(−12) ÷ (2 · 3) = 2
the axis of symmetry
y = 3(4) − 24 + k = k − 12
the height at the vertex
k − 12 = 5
that height is the minimum
k = 17
Range: a > 0, so the vertex is the lowest point and the range is y ≥ 5. When a < 0 the vertex is the highest point and the range is y ≤ the maximum.
Same target, (h, k); three shapes the rule arrives in.
An odd b gives a fraction: y = x² + 3x − 1
x = −3 ÷ (2 · 1) = −32
−b2a
y = (−32)² + 3(−32) − 1
substitute
y = 94 − 184 − 44 = −134
all in quarters
vertex (−32, −134)
a minimum, since a > 0
Nothing changes but the arithmetic: put every term over 4 before adding.
Typing −32² or squaring only the 3 gives −94, and the vertex height comes out −314. Bracket the negative fraction before squaring.
Factored form: y = −(x − 1)(x + 5)
x-intercepts 1 and −5
each bracket = 0
x = (1 + (−5)) ÷ 2 = −2
the vertex is midway
y = −(−3)(3) = 9
put x = −2 into the brackets
vertex (−2, 9), a maximum
the leading minus: a = −1
No expanding needed: the axis is the average of the two roots.
No x term, written backwards: y = 7 − 2x²
a = −2, b = 0, c = 7
the missing x term means b = 0
x = −0 ÷ (−4) = 0, y = 7
−b2a, then substitute
vertex (0, 7), maximum 7
range y ≤ 7
With b = 0 the vertex sits on the y-axis, and a is the number on x² wherever it is written.
Vertex form holds h and k; the extra point fixes the stretch.
y = a(x − 1)² − 8
vertex (1, −8)
−6 = a(0 − 1)² − 8
put in (0, −6)
a = 2
−6 = a − 8
y = 2(x − 1)² − 8
the equation
0 = 2(x − 1)² − 8, so (x − 1)² = 4
x-intercepts: y = 0
x = 3 or x = −1
1 ± 2
y = (x − 1)² − 8 has the right vertex but passes through (0, −7), not (0, −6). The second point is there to fix a — use it.
Vertex items come as stories (highest point, least cost), as a missing constant from a given maximum or minimum, or as a range. Find x = −b2a first, then substitute — and check whether the question wants the x of the vertex (when, where) or the y (how high, the value).
If 20 really is the maximum, nothing nearby can beat it.
A rocket's height after t seconds is
h = −5t² + 30t + 10
Find the maximum height.
t = −b2a = 3 seconds, so h = −5(9) + 90 + 10 = 55 m. C is the launch height at t = 0, and A is the time. That +10 lifts the whole curve, so it has to survive to the last line.
Hint: the + 10 means it was launched from a 10 m platform. It does not affect −b2a at all, but it certainly affects the height.
