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

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,Â **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,Â **the vertex is located at (1, 0).**