5.1 Quadratic Forms, Graphs, Vertex, and Intercepts

5.1 Quadratic Forms, Graphs, Vertex, and Intercepts

College Board’s Next-Generation ACCUPLACER Advanced Algebra and Functions (AAF) test weights quadratics at 10–15% of the exam — typically 2–3 computer-adaptive items out of 20. Table 11 names four skills: creating quadratic equations in one or two variables, solving by factoring or the quadratic formula, simplifying quadratic equations and inequalities, and systems that involve a quadratic. This section is the graph-and-form skill: read a parabola from y = ax^2 + bx + c, convert among the three algebraic forms, and name the vertex, intercepts, and axis of symmetry.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

Domain, range, and extrema of non-quadratic functions — including a downward-opening quartic — belong in Function Notation, Evaluation, Domain, and Range and Graphs of Functions, Vertical Line Test, Maxima and Minima. Stay on parabolas here.

Three forms of a quadratic

A quadratic in one variable has degree 2. The leading coefficient a is not zero. AAF expects three writings of the same parabola.

All three describe the same graph when a is the same number and the constants are consistent. Expanding vertex form recovers standard form. Expanding intercept form also recovers standard form. Completing the square — taught with the quadratic formula in Quadratic Formula and Completing the Square — is the algebraic path from standard to vertex form.

Sign of a: opens up or down

• If a > 0, the parabola opens upward. The vertex is a minimum.
• If a < 0, the parabola opens downward. The vertex is a maximum.
• A large |a| makes a narrow parabola; a small |a| (but not zero) makes a wide one. y = 4x^2 is narrower than y = (1/4)x^2.

AAF items often hide a as a fraction. y = −(1/2)(x − 3)^2 + 7 still opens down because a = −1/2 < 0. The vertex is (3, 7) even though you never expanded the square.

Axis of symmetry and the vertex from standard form

The axis of symmetry is the vertical line through the vertex:

x = −b / (2a)

That formula is not optional. For y = ax^2 + bx + c:

1. Compute h = −b/(2a).
2. Substitute x = h into the equation to get k = y(h).
3. The vertex is (h, k).

The y-intercept is immediate: plug x = 0 to get y = c, so (0, c).

The x-intercepts (roots, zeros) solve ax^2 + bx + c = 0. They may be two distinct reals, a double root, or none. Graphically they are where the parabola crosses or touches the x-axis. Algebraically they are the r1 and r2 in intercept form. If two distinct x-intercepts exist, the vertex x-coordinate is their midpoint: h = (r1 + r2)/2, which matches −b/(2a).

Worked example: y = x^2 − 6x + 5 to vertex form, vertex (3, −4)

Start with standard form y = x^2 − 6x + 5. Here a = 1, b = −6, c = 5.

Axis: x = −(−6)/(2 · 1) = 6/2 = 3.

Vertex y-value: y(3) = 3^2 − 6·3 + 5 = 9 − 18 + 5 = −4.

So the vertex is (3, −4) and vertex form is:

y = (x − 3)^2 − 4

Check by expanding: (x − 3)^2 − 4 = x^2 − 6x + 9 − 4 = x^2 − 6x + 5. The forms match.

Y-intercept: (0, 5), because c = 5. Confirm in vertex form: (0 − 3)^2 − 4 = 9 − 4 = 5.

X-intercepts: factor x^2 − 6x + 5 = (x − 1)(x − 5) = 0, so x = 1 and x = 5. Intercept form is y = (x − 1)(x − 5) with a = 1. The axis sits halfway between the roots: (1 + 5)/2 = 3, matching h = 3.

Range of this parabola: because a = 1 > 0, the graph opens up, the vertex is a minimum, and the range is y ≥ −4, or [−4, ∞) in interval notation. A downward-opening parabola y = −x^2 + … would have range y ≤ k. Do not confuse that with the range of a quartic; AAF graph-to-algebra items in this band are parabolas.

A compact table of values that AAF might encode as a graph:

Points equally far from x = 3 have the same y-value: x = 0 and x = 6 both give y = 5; x = 1 and x = 5 both give y = 0. That is the geometric meaning of the axis of symmetry.

Creating a quadratic in one variable

Table 11 includes creating quadratic equations, not only reading them. A one-variable quadratic often comes from area, a product, or a motion formula h = −16t^2 + v0 t + h0 (height in feet after t seconds).

Area: a rectangle is 3 ft longer than it is wide, and the area is 40 ft². If width is w, then:

w(w + 3) = 40

w^2 + 3w − 40 = 0

That is a quadratic equation in one variable. You will solve it in the next two sections; here the skill is writing it. Do not skip the = 40 and then subtract: the created equation is the product equal to the area, then standard form after moving 40.

Projectile: a ball is thrown upward from 6 ft with initial speed 32 ft/s. Height in feet after t seconds is:

h = −16t^2 + 32t + 6

That is a quadratic function of t. The vertex gives the maximum height; a positive root of h = 0 (if the ground is h = 0) gives landing time. AAF may ask you to create this model from the three numbers 16, 32, and 6 rather than solve it. The coefficient −16 is the feet-and-seconds gravity constant; +32t is the initial-velocity term; +6 is the launch height. Swapping 32 and 6 is a writing error that produces the wrong vertex.

Creating a quadratic in two variables

A two-variable quadratic is a parabola in the plane. Given vertex (h, k) and one other point, write:

y = a(x − h)^2 + k

then solve for a by substituting the extra point.

Worked: vertex (2, −1), and the graph passes through (4, 7).

7 = a(4 − 2)^2 + (−1)

7 = 4a − 1

8 = 4a

a = 2

So y = 2(x − 2)^2 − 1. Expand if the options are in standard form: y = 2(x^2 − 4x + 4) − 1 = 2x^2 − 8x + 8 − 1 = 2x^2 − 8x + 7.

Given x-intercepts r1, r2 and one other point, use intercept form and solve for a.

Worked: x-intercepts at x = −2 and x = 4, and y-intercept (0, −8).

y = a(x + 2)(x − 4)

−8 = a(0 + 2)(0 − 4) = a(2)(−4) = −8a

a = 1

So y = (x + 2)(x − 4) = x^2 − 2x − 8.

If the extra point is not the y-intercept, the same algebra works. Through (1, 9) with roots −1 and 2: 9 = a(1 + 1)(1 − 2) = a(2)(−1) = −2a, so a = −9/2, and y = −(9/2)(x + 1)(x − 2).

Reading a graph without a picture in the stem

CAT items may describe a parabola in words: opens downward, vertex at (1, 5), and passes through (3, −3). Translate immediately:

1. Opens down ⇒ a < 0.
2. Vertex (1, 5) ⇒ y = a(x − 1)^2 + 5.
3. Through (3, −3): −3 = a(2)^2 + 5 ⇒ −8 = 4a ⇒ a = −2.
4. Equation: y = −2(x − 1)^2 + 5.

If options are factored, expand or find the roots from −2(x − 1)^2 + 5 = 0. If options are standard form, expand: y = −2(x^2 − 2x + 1) + 5 = −2x^2 + 4x − 2 + 5 = −2x^2 + 4x + 3.

Vertex versus intercepts — do not swap them

A common wrong choice treats the y-intercept as the vertex. For y = x^2 − 6x + 5 the y-intercept is (0, 5) and the vertex is (3, −4). Another trap reports the axis x = 3 as the point (3, 0) — that point is on the axis only if the vertex happens to sit on the x-axis.

If two x-intercepts exist, the vertex x-coordinate is their midpoint. If there are no x-intercepts, you still have a vertex: use h = −b/(2a). Example: y = x^2 + 4x + 8 has h = −4/2 = −2, k = 4 − 8 + 8 = 4, vertex (−2, 4), and discriminant 16 − 32 = −16 < 0, so it never crosses the x-axis. The graph is still a parabola, and Skills Insight 250–262 still wants that vertex from the algebra.

AAF calculator note for graphs

Some items show an on-screen calculator icon. A decimal vertex such as x = 2.5 is fine when −b/(2a) is a terminating decimal, but AAF often lists exact coordinates such as (3/2, −7/4). Leave fractions as fractions unless the options are decimals. Handheld calculators are not allowed except with an approved accommodation (College Board calculator policy).

Graphing and algebra stay linked through Skills Insight 250–262: if you can move among the three forms and name vertex, intercepts, and opening direction, you have the first quadratic skill. Solving by factoring is next, at Solving Quadratics by Factoring.