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Quadratic Function From a Table of Values

The table shows values for a quadratic function. What is the axis of symmetry and an explicit formula for the function?

Subject: Math Algebra Chapter: Functions Topic: Discontinuity Classifier Answer included
the table shows values for a quadratic function quadratic function table finite differences second differences polynomial interpolation axis of symmetry vertex form standard form
Accepted answer Answer included

The table shows values for a quadratic function. The constant second differences confirm a quadratic model, and the repeated y-values at equal distances from a central x-value reveal the axis of symmetry.

Given values

Let the function be written as f of x. The data below use equally spaced x-values.

x f(x)
−16
01
1−2
2−3
3−2
41
56

Symmetry and finite differences

Equal y-values appear in pairs around : , , . This symmetry indicates an axis of symmetry at .

Forward differences are written with . With unit spacing in x, \( \Delta f(x)=f(x+1)-f(x) \) and \( \Delta^2 f(x)=\Delta f(x+1)-\Delta f(x) \).

x f(x) Δf Δ²f
−16−52
01−32
1−2−12
2−312
3−232
415
56

The constant value \( \Delta^2 f = 2 \) confirms a quadratic function (a parabola) and pins down the leading coefficient. For \( f(x)=a x^2+b x+c \) with unit spacing, \[ \Delta f(x)=a\big((x+1)^2-x^2\big)+b\big((x+1)-x\big)=a(2x+1)+b, \] \[ \Delta^2 f(x)=\big(a(2(x+1)+1)+b\big)-\big(a(2x+1)+b\big)=2a. \] With \( \Delta^2 f = 2 \), the coefficient is \( a=1 \).

Function rule (standard form and vertex form)

With \( a=1 \), the model is \( f(x)=x^2+b x+c \). From the table, \( f(0)=1 \) gives \( c=1 \). Also \( f(1)=-2 \) gives \[ -2 = 1^2 + b\cdot 1 + 1 \quad \Rightarrow \quad b = -4. \] The quadratic function is \[ f(x)=x^2-4x+1. \]

Vertex form follows by completing the square: \[ x^2-4x+1=(x^2-4x+4)-3=(x-2)^2-3. \] The axis of symmetry is \( x=2 \), matching the symmetry visible in the table.

y = (x − 2)² − 3 x y
The curve and points represent the same quadratic function; the dashed vertical line marks the axis of symmetry \(x=2\), and the lowest point is the vertex \((2,-3)\).

Key features implied by the table and the formula

  • Axis of symmetry: \(x=2\).
  • Vertex: \((2,-3)\), from \(f(x)=(x-2)^2-3\).
  • y-intercept: \((0,1)\), from \(f(0)=1\).
  • x-intercepts: solutions of \(x^2-4x+1=0\), \[ x=\frac{4\pm\sqrt{16-4}}{2}=\frac{4\pm\sqrt{12}}{2}=2\pm\sqrt{3}\approx 0.268,\; 3.732. \]
  • Opening direction and minimum value: \(a=1>0\) opens upward, minimum \(f(2)=-3\).
  • Domain and range: domain \( (-\infty,\infty) \), range \( [-3,\infty) \).

For equally spaced x-values with step \(h\), a quadratic \(f(x)=a x^2+b x+c\) has constant second differences \[ \Delta^2 f = 2a h^2. \] With \(a\) fixed by \( \Delta^2 f \), any two table points determine \(b\) and \(c\), yielding the unique quadratic interpolant consistent with the table.

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