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Absolute lowest point of a quadratic function

absolute lowest point: For the quadratic function \(f(x)=x^2-4x+1\), find the absolute lowest point (global minimum) and the minimum value, then state the range of \(f\) over all real \(x\).

Subject: Math Algebra Chapter: Functions Topic: Domain and Range Calculator Answer included
absolute lowest point absolute minimum global minimum vertex of a parabola completing the square vertex form minimum value axis of symmetry
Accepted answer Answer included

The phrase absolute lowest point refers to the point on the graph where the function attains its smallest possible \(y\)-value over its domain (a global minimum). For a quadratic function with a positive leading coefficient, that point is the vertex of the parabola.

\[ f(x)=x^2-4x+1. \]

Vertex form and the absolute lowest point

Completing the square rewrites the quadratic so the minimum is visible directly:

\[ f(x)=x^2-4x+1 =\bigl(x^2-4x+4\bigr)-4+1 =(x-2)^2-3. \]

Since \((x-2)^2\ge 0\) for every real \(x\), the smallest possible value of \((x-2)^2-3\) occurs when \((x-2)^2=0\), which happens at \(x=2\).

Minimum value and range

\[ f(2)=(2-2)^2-3=-3. \]

The absolute lowest point is therefore \(\,(2,-3)\,\). The minimum value is \(-3\), and the range over all real numbers is

\[ y\ge -3. \]

Summary table

Item Result Meaning
Vertex form \(f(x)=(x-2)^2-3\) Shows the minimum shift from a perfect square
Axis of symmetry \(x=2\) Vertical line through the vertex
Absolute lowest point \((2,-3)\) Global minimum point of the parabola
Minimum value \(-3\) Smallest possible output
Range \(y\ge -3\) All attainable outputs

Visualization

Graph of f(x)=x²−4x+1 with the absolute lowest point marked Coordinate axes with a parabola opening upward. The vertex at (2, −3) is highlighted as the absolute lowest point. The axis of symmetry x=2 is shown as a dashed vertical line. −1 0 1 2 3 4 5 −4 −2 0 2 4 6 8 x y absolute lowest point: (2, −3) axis of symmetry: x = 2 2 − √3 2 + √3 f(x) = x² − 4x + 1
The vertex marks the absolute lowest point \((2,-3)\) because the parabola opens upward, and the squared term in \((x-2)^2-3\) cannot be negative.

Common pitfalls

The y-intercept is not generally the absolute lowest point. The sign of the leading coefficient controls the opening direction: a positive coefficient produces a global minimum (absolute lowest point), while a negative coefficient produces a global maximum.

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