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Find the coordinates of the vertex for the parabola defined by the given quadratic function.

\[ \boldsymbol{ f ( x ) \ = \ 2 x^{ 2 } \ – \ 8 x \ + \ 3 } \]

The aim of this question is to learn how to evaluate the vertex location of a parabola.

A U-shaped curve that follows the quadratic law (its equation is quadratic), is called a parabola. A parabola has a mirror like symmetry. The point on a parabolic curve that touches its symmetrical axis is called a vertex. Given a parabola of the form:

\[ f ( x ) \ = \ a x^{ 2 } \ + \ b x \ + \ c \]

The x-coordinate of its vertex can be evaluated by using the following formula:

\[ h \ = \ \dfrac{ – b }{ 2a } \]

Expert Answer

Given that:

\[ f ( x ) \ = \ 2 x^{ 2 } \ – \ 8 x \ + \ 3 \]

Comparing with the standard form of quadratic equation, we can conclude that:

\[ a \ = \ 2 \]

\[ b \ = \ -8 \]

\[ c \ = \ 3 \]

Recall the standard formula for the x-coordinate of the vertex of a parabola:

\[ h \ = \ \dfrac{ – b }{ 2a } \]

Substituting values:

\[ h \ = \ \dfrac{ – ( -8 ) }{ 2 ( 2 ) } \]

\[ \Rightarrow h \ = \ \dfrac{ 8 }{ 4 } \]

\[ \Rightarrow h \ = \ 2 \]

To find the y-coordinate, we simply evaluate the given equation of the parabola at x = 2. Recall:

\[ f ( x ) \ = \ 2 x^{ 2 } \ – \ 8 x \ + \ 3 \]

Substituting x = 2 in the above equation:

\[ f ( 2 ) \ = \ 2 ( 2 )^{ 2 } \ – \ 8 ( 2 ) \ + \ 3 \]

\[ \Rightarrow f ( 2 ) \ = \ 2 ( 4 ) \ – \ 8 ( 2 ) \ + \ 3 \]

\[ \Rightarrow f( 2 ) \ = \ 8 \ – \ 16 \ + \ 3 \]

\[ \Rightarrow f ( 2 ) \ = \ -5 \]

Hence, the vertex is located at (2, -5).

Numerical Result

The vertex is located at (2, -5).

Example

Given the following equation of a parabola, find the location of its vertex.

\[ \boldsymbol{ f ( x ) \ = \ x^{ 2 } \ – \ 2 x \ + \ 1 } \]

For the x-coordinate of the vertex:

\[ h \ = \ \dfrac{ – ( -2 ) }{ 2 ( 1 ) } \]

\[ \Rightarrow h \ = \ \dfrac{ 2 }{ 2 } \]

\[ \Rightarrow h \ = \ 1 \]

To find the y-coordinate, we simply evaluate the given equation of the parabola at x = 1. Recall:

\[ f ( 2 ) \ = \ ( 1 )^{ 2 } \ – \ 2 ( 1 ) \ + \ 1 \]

\[ \Rightarrow f( 2 ) \ = \ 1 \ – \ 2 \ + \ 1 \]

\[ \Rightarrow f ( 2 ) \ = \ 0 \]

Hence, the vertex is located at (1, 0).

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