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VirginiaMathsSyllabus dot point

How do you graph and analyze a quadratic function, including its vertex, axis of symmetry, and the forms of its equation?

Graph and analyze quadratic functions, identifying the vertex, axis of symmetry, intercepts, and direction of opening, and connecting standard, vertex, and factored forms (A.F.5).

A Virginia SOL Algebra I answer on A.F.5: the parabola, finding the vertex and axis of symmetry, direction of opening, the three forms of a quadratic, and reading intercepts.

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  1. What this topic is asking
  2. The parabola and its direction
  3. The axis of symmetry and the vertex
  4. The three forms
  5. Reading intercepts
  6. Why every parabola is symmetric about the vertex
  7. How the SOL examines this topic
  8. Try this

What this topic is asking

A.F.5 asks you to graph and analyze quadratic functions: find the vertex, axis of symmetry, intercepts, and direction of opening, and connect the three forms of the equation. On the Virginia Algebra I SOL these are Functions items: identify the vertex or axis, decide which way a parabola opens, or read features from a form. They appear as multiple choice, fill-in-the-blank, and coordinate-plane items.

The parabola and its direction

A quadratic function f(x)=ax2+bx+cf(x) = ax^2 + bx + c graphs as a parabola, a symmetric U-shaped curve. The leading coefficient aa controls two things:

  • Direction: a>0a > 0 opens upward (a minimum at the vertex); a<0a < 0 opens downward (a maximum at the vertex).
  • Width: a larger ∣a∣|a| makes a narrower parabola; a smaller ∣a∣|a| makes a wider one.

The axis of symmetry and the vertex

The axis of symmetry is the vertical line that splits the parabola into mirror halves. It passes through the vertex, and its equation is

x=βˆ’b2a.x = \frac{-b}{2a}.

This formula is not on the Algebra I formula sheet, so memorize it. The vertex is the turning point; its xx-coordinate is the axis value, and its yy-coordinate is found by evaluating the function there.

The three forms

The same parabola can be written three ways, each displaying a different feature:

  • Standard form f(x)=ax2+bx+cf(x) = ax^2 + bx + c: the constant cc is the yy-intercept, and aa shows the direction.
  • Vertex form f(x)=a(xβˆ’h)2+kf(x) = a(x - h)^2 + k: the vertex is (h,k)(h, k) directly (watch the sign: (xβˆ’3)(x - 3) means h=3h = 3).
  • Factored form f(x)=a(xβˆ’r1)(xβˆ’r2)f(x) = a(x - r_1)(x - r_2): the zeros are r1r_1 and r2r_2.

Completing the square converts standard to vertex form; factoring converts standard to factored form.

Reading intercepts

The yy-intercept is f(0)=cf(0) = c in standard form. The xx-intercepts (zeros) come from setting f(x)=0f(x) = 0 and solving (factoring, square roots, or the quadratic formula). A parabola can have two xx-intercepts, one (vertex on the axis, a double root), or none (it never reaches the xx-axis), matching the discriminant.

Why every parabola is symmetric about the vertex

The symmetry of a parabola, and the axis formula, both come from the structure of a quadratic. Two inputs that are equally far from the axis produce the same output, because the squared term treats equal distances on either side identically: a(xβˆ’h)2a(x - h)^2 gives the same value for h+dh + d and hβˆ’dh - d. That mirror symmetry is why the parabola folds onto itself along the line x=hx = h, the axis of symmetry, and why the vertex sits exactly at the fold, where the two halves meet. The formula x=βˆ’b2ax = \frac{-b}{2a} is just the axis location written in terms of the standard-form coefficients (it is the hh you would get by completing the square). This symmetry is practically useful: once you know the vertex and one other point, you automatically know its mirror image, so you can sketch a parabola from very little, and it explains why a parabola's two zeros are always equidistant from the axis.

How the SOL examines this topic

  • Multiple choice. Identify the direction of opening, the vertex, or the axis of symmetry.
  • Fill-in-the-blank. Compute the axis of symmetry or the vertex coordinates.
  • Coordinate-plane items. Plot the vertex, or match a parabola to its equation form.

Try this

Q1. Find the axis of symmetry of f(x)=x2+4xβˆ’1f(x) = x^2 + 4x - 1. [1 point]

  • Cue. x=βˆ’42(1)=βˆ’2x = \frac{-4}{2(1)} = -2.

Q2. Does f(x)=3x2βˆ’5f(x) = 3x^2 - 5 open up or down? [1 point]

  • Cue. a=3>0a = 3 > 0, so up (a minimum).

Exam-style practice questions

Practice questions written in the style of VDOE exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SOL (style)2 marksFill in the blank. Find the axis of symmetry and the vertex of f(x)=x2βˆ’6x+5f(x) = x^2 - 6x + 5.
Show worked answer β†’

The axis of symmetry is x=3x = 3 and the vertex is (3,βˆ’4)(3, -4).

The axis of symmetry is x=βˆ’b2a=βˆ’(βˆ’6)2(1)=62=3x = \frac{-b}{2a} = \frac{-(-6)}{2(1)} = \frac{6}{2} = 3. Substitute x=3x = 3 to find the vertex's yy-value: f(3)=9βˆ’18+5=βˆ’4f(3) = 9 - 18 + 5 = -4, so the vertex is (3,βˆ’4)(3, -4). The axis formula is not on the sheet, so it must be memorized; forgetting the sign of βˆ’b-b is the common slip.

SOL (style)1 marksMultiple choice. For the function f(x)=βˆ’2x2+4x+1f(x) = -2x^2 + 4x + 1, which way does the parabola open and what feature is at the vertex? (A) downward, a maximum (B) upward, a minimum (C) downward, a minimum (D) upward, a maximum
Show worked answer β†’

The correct answer is (A).

The leading coefficient a=βˆ’2a = -2 is negative, so the parabola opens downward. A downward parabola has its highest point at the vertex, so the vertex is a maximum. A positive aa would open upward with a minimum at the vertex. The sign of aa alone decides the direction of opening.

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