Five things worth knowing about any parabola.
| Feature | How to find it |
|---|---|
| Opening | a > 0 up, a < 0 down |
| Axis of symmetry | x = −b2a |
| Vertex | that x, then substitute for y |
| y-intercept | (0, c) |
| x-intercepts | solve y = 0 |
- Home
- Grade 7-12 Math Review
- Quadratic Functions and Their Graphs
Quadratic Functions and Their Graphs
Vertex, axis, intercepts — four or five points and the curve draws itself.
Find the vertex
What is the vertex of y = x² − 8x + 11?
The axis is at x = −b2a = −−82 = 4. Substituting: y = 16 − 32 + 11 = −5, so the vertex is (4, −5). B mishandles the double negative — b is −8, so −b is +8, not −8. C reads the y-value off the constant term, but 11 is the y-intercept, the height at x = 0, not at the vertex.
Graphing, in order.
y = x² − 8x + 11
a = 1, opens up
axis: x = 4, vertex (4, −5)
the lowest point
y-intercept (0, 11)
the constant
mirror it: (8, 11)
same distance the other side of x = 4
Symmetry gives you a free point. Anything 4 units left of the axis has a twin 4 units right of it, so every point you compute is really two.
What a controls.
a = 1
the standard parabola
opens up
a = 4
four times as steep
narrower
a = 14
a quarter as steep
wider
a = −1
reflected in the x-axis (flipped upside down)
opens down
The sign of a decides the direction; the size decides the width. A larger |a| means a narrower curve, which catches people out — bigger number, thinner shape.
Converting to vertex form.
y = x² − 8x + 11
half of −8 is −4, squared is 16
= (x² − 8x + 16) + 11 − 16
add and subtract 16
= (x − 4)² − 5
vertex (4, −5) ✓
Same vertex as the −b2a route, which is a useful cross-check. Vertex form is worth the extra work when you also need to describe the transformations.
Reading a graph from its form: shifts, vertex, range, axis.
y = a(x − h)² + k: vertex (h, k), axis x = h, and k is the maximum or minimum.
Describe y = −2(x + 4)² + 3 and give its range.
x + 4 = x − (−4), so h = −4
shifted 4 to the LEFT
k = 3
shifted 3 up
a = −2
opens down, narrower than y = x²
vertex (−4, 3), range y ≤ 3
3 is the maximum
The sign inside the bracket is the opposite of the shift
Reading (x + 4)² as "h = 4" puts the vertex at (4, 3) — on the wrong side of the y-axis. Test it: the square is zero when x + 4 = 0, that is at x = −4, so the vertex is (−4, 3).
The x-intercepts are −2 and 6. Find the axis and the vertex of y = (x + 2)(x − 6).
axis: x = (−2 + 6) ÷ 2 = 2
midway between the intercepts
y = (2 + 2)(2 − 6) = 4 × (−4) = −16
substitute the axis value
vertex (2, −16)
a = 1 > 0, so a minimum
The two intercepts are mirror images across the axis, so the axis is always their average.
Variations: a not 1, a negative a, no x term.
The same vertex method in the three forms that most often trip it.
a is not 1: write y = 2x² − 8x + 3 in vertex form.
= 2(x² − 4x) + 3
factor 2 from the x terms only
= 2(x² − 4x + 4) − 8 + 3
the 4 inside is worth 2 × 4 = 8
= 2(x − 2)² − 5
check: −b2a = 84 = 2 ✓
Different: what we add inside the bracket is multiplied by a, so we take away a times it.
A negative a: the maximum of y = −x² + 4x + 1
x = −42 × (−1) = 2
the axis
y = −4 + 8 + 1 = 5
substitute x = 2
maximum 5, range y ≤ 5
a < 0 opens down
Different: with a < 0 the vertex is the highest point, so the range points downward from k.
No x term: y = 3x² − 12
b = 0, so the axis is x = 0
the y-axis itself
vertex (0, −12)
also the y-intercept
3x² = 12, x² = 4, x = ±2
x-intercepts, mirror images
Different: with b = 0 the vertex and the y-intercept are the same point.
How many x-intercepts, and where.
x² − 8x + 11 = 0
D = 64 − 44 = 20
x = 8 ± √202 = 4 ± √5
≈ 1.76 and 6.24
Vertex below the axis, opens up
two crossings
D > 0
Vertex on the axis
one touch
D = 0
Vertex above the axis, opens up
none
D < 0
Note the two intercepts are 4 ± √5, symmetric about the axis x = 4 — as they must be.
On the exam
Items ask for the vertex, the axis, the maximum or minimum value, the range, or which way a graph has moved from y = x². The fastest route is x = −b2a and substitute, or read (h, k) straight off vertex form. The 10-second check: the y-intercept c and its mirror point across the axis must have the same height.
Now you try
Practice problem
Find the minimum
What is the minimum value of y = 2x² + 12x + 7?
The axis is x = −122 × 2 = −3. Substituting: y = 2(9) − 36 + 7 = 18 − 36 + 7 = −11. Since a = 2 is positive the parabola opens up, so this is a minimum. A is where the minimum occurs, not its value. B is the y-intercept.
Hint: find the x of the vertex first, then substitute it back. The value is always the y-coordinate.