Skip to main content
TennesseeMathsSyllabus dot point

How do you graph a quadratic function and identify its vertex, axis of symmetry, intercepts, and direction of opening?

Graph quadratic functions and show key features including the vertex, axis of symmetry, intercepts, maximum or minimum, and direction of opening (TN A1.F.IF.D.7a).

A TNReady Algebra I answer on graphing quadratics (TN A1.F.IF.D.7a), finding the vertex with the axis of symmetry, the y-intercept and x-intercepts, the direction of opening, and reading maximum or minimum.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this topic is asking
  2. The features and how to find them
  3. Symmetry makes graphing fast
  4. How TNReady examines this topic
  5. Why the sign of a decides so much
  6. Try this

What this topic is asking

Standard A1.F.IF.D.7a asks you to graph a quadratic and show its key features: the vertex, the axis of symmetry, the intercepts, whether it has a maximum or minimum, and its direction of opening. A parabola is symmetric about a vertical line through its vertex, and reading these features (by hand in simple cases) is a core Functions-category skill that overlaps the Quadratic strand.

The features and how to find them

Feature How to find it
Direction of opening sign of aa: up if positive, down if negative
Max or min opens up gives a minimum; opens down gives a maximum
Axis of symmetry x=βˆ’b2ax = \frac{-b}{2a} (a vertical line)
Vertex (βˆ’b2a,Β f(βˆ’b2a))\left(\frac{-b}{2a},\ f\left(\frac{-b}{2a}\right)\right)
yy-intercept cc (evaluate at x=0x = 0)
xx-intercepts (zeros) solve ax2+bx+c=0ax^2 + bx + c = 0

Symmetry makes graphing fast

Because a parabola is symmetric about its axis, every point has a mirror image the same distance on the other side. Once you have the vertex and one extra point (often the yy-intercept), reflecting gives a third point for free, enough to sketch the curve. This is also why the two xx-intercepts are equidistant from the axis, so the axis sits exactly halfway between them.

How TNReady examines this topic

  • Numeric response. Find the axis, a vertex coordinate, or an intercept.
  • Multiple choice. Identify the direction of opening, max or min, or match a graph to an equation.
  • Graphing. Plot the vertex and intercepts of a parabola.

A clarifying idea is that the zeros from the solving topic are precisely the xx-intercepts here, and the vertex from completing the square is the same vertex you locate with x=βˆ’b2ax = \frac{-b}{2a}. The solving and graphing strands describe one object from two angles.

Why the sign of a decides so much

The leading coefficient aa controls the parabola's shape and orientation before any other feature. A positive aa opens the curve upward into a valley, so the vertex is the lowest point (a minimum) and the function decreases then increases. A negative aa flips it into a hill, so the vertex is the highest point (a maximum) and the function increases then decreases. The magnitude of aa also sets how narrow or wide the parabola is, a larger ∣a∣|a| makes it steeper. On a multiple-choice item that asks for direction and max-or-min, you do not need the vertex at all: the single sign of aa answers it, which is the fastest possible route to the point. This is why checking the sign of aa is always the first move when you meet a quadratic graph.

Try this

Q1. For f(x)=x2+2xβˆ’8f(x) = x^2 + 2x - 8, find the axis of symmetry and the vertex. [2 points]

  • Cue. Axis x=βˆ’22=βˆ’1x = \frac{-2}{2} = -1; f(βˆ’1)=1βˆ’2βˆ’8=βˆ’9f(-1) = 1 - 2 - 8 = -9, vertex (βˆ’1,βˆ’9)(-1, -9).

Q2. Does f(x)=βˆ’3x2+6xf(x) = -3x^2 + 6x open up or down, and is the vertex a max or min? [1 point]

  • Cue. a=βˆ’3<0a = -3 < 0: opens down, vertex is a maximum.

Exam-style practice questions

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

TNReady (style)2 marksNumeric response. For f(x)=x2βˆ’6x+5f(x) = x^2 - 6x + 5, find the x-coordinate of the vertex.
Show worked answer β†’

The vertex's xx-coordinate is x=3x = 3.

The axis of symmetry is x=βˆ’b2ax = \frac{-b}{2a}. With a=1a = 1 and b=βˆ’6b = -6: x=βˆ’(βˆ’6)2(1)=62=3x = \frac{-(-6)}{2(1)} = \frac{6}{2} = 3. The vertex lies on this vertical line. (Its yy-coordinate is f(3)=9βˆ’18+5=βˆ’4f(3) = 9 - 18 + 5 = -4, so the vertex is (3,βˆ’4)(3, -4).) Using x=βˆ’b2ax = \frac{-b}{2a} to locate the axis and the vertex is the central skill for graphing a parabola.

TNReady (style)2 marksMultiple choice. The graph of f(x)=βˆ’2x2+8xβˆ’3f(x) = -2x^2 + 8x - 3 opens in which direction and has what kind of vertex? (A) opens down, maximum (B) opens down, minimum (C) opens up, maximum (D) opens up, minimum
Show worked answer β†’

The correct answer is (A).

The leading coefficient is a=βˆ’2a = -2. A negative aa means the parabola opens downward, so its vertex is the highest point, a maximum. A positive aa would open upward with a minimum. The sign of aa alone settles both the direction of opening and whether the vertex is a max or min, before any other computation.

Related dot points

Sources & how we know this