6.1 Key Features of Quadratic Functions: Intercepts, Vertex, Symmetry & Concavity
Key Takeaways
- For f(x) = ax² + bx + c the axis of symmetry is x = −b/(2a), and the vertex is the point on the parabola at that x-value.
- Concavity is set by the sign of a: a > 0 opens upward (concave up) with a minimum at the vertex, and a < 0 opens downward with a maximum.
- The y-intercept of ax² + bx + c is always (0, c), readable without any computation.
- The x-intercepts, also called zeros or roots, are symmetric about the axis of symmetry, so their average equals the vertex x-coordinate.
- In projectile contexts the vertex gives the maximum height and the time it occurs, while the positive x-intercept gives the time of landing.
6.1 Key Features of Quadratic Functions
Skill 7 of Competency 2 names five features explicitly — x-intercepts, y-intercept, vertex, line of symmetry, and concavity — and requires you to interpret them, not merely compute them. Because the test provides no graphing calculator, every one of these must be found algebraically.
The three forms of a quadratic
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| Standard f(x) = ax^2 + bx + c reveals: y-intercept (0, c) |
| Vertex f(x) = a(x - h)^2 + k reveals: vertex (h, k) |
| Factored f(x) = a(x - r)(x - s) reveals: x-intercepts r and s |
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Each form hands you one feature for free. Recognizing which form you have tells you which feature costs nothing.
Concavity
The sign of the leading coefficient a controls the opening direction:
- a > 0 → concave up, the parabola opens upward, the vertex is the minimum, and the range is y ≥ k.
- a < 0 → concave down, the parabola opens downward, the vertex is the maximum, and the range is y ≤ k.
The magnitude |a| controls width: |a| > 1 makes a narrow parabola, and 0 < |a| < 1 makes a wide one. Note that a can never be 0 — that would make the function linear.
The axis of symmetry and the vertex
Every parabola is symmetric about a vertical line through its vertex, the line of symmetry (or axis of symmetry).
Axis of symmetry: x = −b/(2a)
Find the vertex by computing that x-value, then substituting back to get y.
f(x) = 2x² − 12x + 5 Axis: x = −(−12)/(2 · 2) = 12/4 = 3 Vertex y: f(3) = 2(9) − 12(3) + 5 = 18 − 36 + 5 = −13 Vertex: (3, −13), a minimum since a = 2 > 0. Line of symmetry: x = 3.
Two sign errors dominate here. First, the formula is −b/(2a), so a negative b produces a positive axis value. Second, the denominator is 2a, not a; forgetting the 2 halves nothing and doubles the answer.
From vertex form the work is free: f(x) = −3(x + 4)² + 7 has vertex (−4, 7), axis x = −4, opens downward, maximum value 7. Remember the sign flip: (x + 4) means h = −4.
Intercepts
The y-intercept is found by setting x = 0. In standard form this is instantly (0, c). For f(x) = 2x² − 12x + 5, the y-intercept is (0, 5). From vertex form you must substitute: f(0) = −3(0 + 4)² + 7 = −48 + 7 = −41.
The x-intercepts (zeros, roots, solutions) are found by setting f(x) = 0. From factored form they are immediate: f(x) = 2(x − 3)(x + 5) has zeros at x = 3 and x = −5. Otherwise you must factor, complete the square, or apply the quadratic formula — the subject of the next two sections.
A parabola may have two, one, or zero x-intercepts, depending on whether the vertex sits below, on, or above the x-axis relative to the opening direction.
Symmetry as a computational shortcut
The zeros are equidistant from the axis of symmetry, so:
axis of symmetry = (r + s)/2, the average of the zeros
If a parabola crosses at x = −1 and x = 7, its axis is x = (−1 + 7)/2 = 3, and the vertex is at (3, f(3)) — no formula needed. This works in reverse too: given a zero at x = 2 and a vertex at x = 5, the other zero is at x = 8, since 5 is the midpoint.
Symmetry also relates ordinary function values. Because f(1) and f(5) sit equally far from an axis at x = 3, they are equal. Items test this by giving a partial table and asking for a missing value.
Interpreting features in context
Projectile motion. A ball's height in feet after t seconds is h(t) = −16t² + 64t + 5.
- Concavity: a = −16 < 0, so the path opens downward — the ball rises then falls. The vertex is a maximum.
- Axis / vertex: t = −64/(2 · −16) = 64/32 = 2 seconds. Then h(2) = −16(4) + 64(2) + 5 = −64 + 128 + 5 = 69 feet. The ball reaches a maximum height of 69 feet at t = 2 s.
- y-intercept: h(0) = 5, the launch height — the ball left the hand 5 feet above the ground.
- x-intercept: the positive solution of −16t² + 64t + 5 = 0 gives the time the ball hits the ground.
Note that the negative x-intercept is mathematically valid but meaningless in context, since negative time precedes the launch. Items commonly ask which root is contextually appropriate.
Revenue. If revenue is R(p) = −20p² + 700p, the vertex gives the price maximizing revenue: p = −700/(2 · −20) = 700/40 = $17.50, with maximum revenue R(17.50) = −20(306.25) + 700(17.50) = −6,125 + 12,250 = $6,125. The zeros at p = 0 and p = 35 are the prices producing no revenue — giving it away, or charging so much that nobody buys.
Area. For a rectangle with a fixed 80-foot perimeter, width w gives length 40 − w and area A(w) = w(40 − w) = −w² + 40w. The vertex at w = 20 shows that the square maximizes area, a result worth remembering.
For f(x) = −4x² + 24x − 11, identify the line of symmetry and state whether the vertex is a maximum or a minimum.
A quadratic function has x-intercepts at x = −5 and x = 9. What is the x-coordinate of its vertex?
A rocket's height in meters is h(t) = −5t² + 40t + 12, where t is seconds after launch. What does the y-intercept represent, and what is the maximum height?