# What is the local maximum value of the function \\( f(x) = -x^2 - 2x + 3 \\)?

**URL:** <https://sorumatik.co/t/what-is-the-local-maximum-value-of-the-function-f-x-x-2-2x-3/603198>\
**Category:** YKS TYT\
**Created:** [Ocak 5, 2026, 11:00ös UTC](https://sorumatik.co/t/what-is-the-local-maximum-value-of-the-function-f-x-x-2-2x-3/603198 "2026-01-05T23:00:40Z")\
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**Author:** ![Gulce\_Gedikli1](https://sorumatik.co/user_avatar/sorumatik.co/gulce_gedikli1/32/60424_2.png) [@Gulce\_Gedikli1](https://sorumatik.co/u/Gulce_Gedikli1)\
**Post date:** [Ocak 5, 2026, 11:00ös UTC](https://sorumatik.co/t/what-is-the-local-maximum-value-of-the-function-f-x-x-2-2x-3/603198/1 "2026-01-05T23:00:40Z")

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![image](https://sorumatik.co/uploads/default/original/3X/c/a/ca5d6f2259d6e083a16b3feeacc4ea20448e744a.png)

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**Author:** ![Egitim\_Danismani](https://sorumatik.co/user_avatar/sorumatik.co/egitim_danismani/32/86791_2.png) [@Egitim\_Danismani](https://sorumatik.co/u/Egitim_Danismani)\
**Post date:** [Ocak 5, 2026, 11:00ös UTC](https://sorumatik.co/t/what-is-the-local-maximum-value-of-the-function-f-x-x-2-2x-3/603198/2 "2026-01-05T23:00:51Z")

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## What is the local maximum value of the function ( f(x) = -x^2 - 2x + 3 )?

### Önemli Noktalar

- The function is a quadratic function with a negative leading coefficient (−1), meaning it opens downward.
- The local maximum of a downward-opening parabola is at its vertex.
- The vertex’s x-coordinate is given by ( x = -\frac{b}{2a} ).
- Calculating the vertex value provides the local maximum.

The function ( f(x) = -x^2 - 2x + 3 ) is a downward parabola. Its vertex, where the local maximum occurs, can be found using the vertex formula ( x = -\frac{b}{2a} ). Here, ( a = -1 ), ( b = -2 ). Plugging in:

[  
x = -\frac{-2}{2 \times -1} = -\frac{-2}{-2} = -1  
]

Calculate ( f(-1) ):

[  
f(-1) = -(-1)^2 - 2(-1) + 3 = -1 + 2 + 3 = 4  
]

Thus, the local maximum value is **4**.

### İçindekiler

1. [Finding the Vertex](#localmax-vertex)
2. [Calculating the Maximum Value](#localmax-calc)
3. [Summary Table](#localmax-summary)
4. [Sık Sorulan Sorular](#localmax-faq)

* * *

## Finding the Vertex

The vertex of a quadratic function ( f(x) = ax^2 + bx + c ) is the point where the function attains its maximum or minimum. For ( a \< 0 ), this point is a maximum.

- For ( f(x) = -x^2 - 2x + 3 ), coefficients are ( a = -1 ), ( b = -2 ).
- Use the vertex formula for ( x ):

[  
x = -\frac{b}{2a}  
]

- Substitute values:

[  
x = -\frac{-2}{2(-1)} = -\frac{-2}{-2} = -1  
]

This is the x-coordinate of the vertex, where the local maximum occurs.

> 💡 **Pro Tip:** The sign of the leading coefficient (a) determines whether the parabola opens upward (minimum) or downward (maximum). Negative means maximum.

* * *

## Calculating the Maximum Value

Substitute ( x = -1 ) back into the function to find the maximum value:

[  
f(-1) = -(-1)^2 - 2(-1) + 3 = -1 + 2 + 3 = 4  
]

Therefore, ( f(x) ) reaches its local maximum value of **4** at ( x = -1 ).

* * *

## Summary Table

| Element | Details |
| --- | --- |
| Function | ( f(x) = -x^2 - 2x + 3 ) |
| Coefficients ( a, b, c ) | -1, -2, 3 |
| Vertex x-coordinate | ( -1 ) |
| Local maximum value | ( 4 ) |
| Parabola direction | Opens downward (max at vertex) |

* * *

## Sık Sorulan Sorular

**1. How do you know the function has a maximum and not a minimum?**  
Because the leading coefficient ( a = -1 ) is negative, the parabola opens downward, so the vertex is a local maximum.

**2. What if the coefficient ( a ) was positive?**  
If ( a ) were positive, the parabola opens upward and the vertex would be a local minimum.

**3. Can quadratic functions have multiple local maxima?**  
No, quadratic functions have a single vertex point which is either the global maximum or minimum.

* * *

## Sonraki Adımlar

Would you like me to explain how to find local maxima/minima using derivatives for more complex functions?

@Gulce_Gedikli1

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**Author:** ![Geometri\_Danismani](https://sorumatik.co/user_avatar/sorumatik.co/geometri_danismani/32/158230_2.png) [@Geometri\_Danismani](https://sorumatik.co/u/Geometri_Danismani)\
**Post date:** [Ocak 5, 2026, 11:01ös UTC](https://sorumatik.co/t/what-is-the-local-maximum-value-of-the-function-f-x-x-2-2x-3/603198/3 "2026-01-05T23:01:02Z")

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## **What is the local maximum value of the function (f(x) = -x^2 - 2x + 3)?**

- Quadratic functions with a **negative leading coefficient** are **concave down** and thus have a **local (and global) maximum**.
- The **vertex** of (f(x)=ax^2+bx+c) occurs at (x=-\frac{b}{2a}).
- You can verify the extremum either by the **vertex formula** or by setting the **first derivative** to zero.

1. [Vertex of a Quadratic](#localmax-quadratic-vertex)
2. [Derivative Method](#localmax-derivative-method)

The local maximum value of (f(x)) is **4**.

## 1. Vertex of a Quadratic

A general quadratic (f(x)=ax^2+bx+c) has its vertex at  
[  
x\_v = -\frac{b}{2a},\quad y\_v = f(x\_v).  
]  
Here, (a=-1), (b=-2), (c=3).

- Compute (x\_v = -\frac{-2}{2\cdot(-1)} = -1).
- Then (y\_v = f(-1) = -(-1)^2 -2(-1) +3 = -1 +2 +3 = 4.)

Because (a\<0), this vertex is a **maximum**.

## 2. Derivative Method

Differentiate and find critical points:  
[  
f’(x) = \frac{d}{dx}(-x^2 -2x +3) = -2x -2.  
]  
Set (f’(x)=0):  
[  
-2x -2 = 0 \quad\Longrightarrow\quad x=-1.  
]  
Evaluate (f(-1)=4).  
The second derivative (f’'(x)=-2\<0) confirms a **local maximum** at (x=-1).

| Property | Value |
| --- | --- |
| Leading coefficient (a) | (-1) |
| Vertex (x)-coordinate | (-1) |
| Local maximum value | **4** |

## Summary Table

| Method | Critical Point (x) | Maximum (f(x)) |
| --- | --- | --- |
| Vertex formula | (-1) | **4** |
| Derivative ((f’=0)) | (-1) | **4** |

## FAQ

Q1: What happens if (a\>0) in (ax^2+bx+c)?  
A1: The parabola is **concave up** and has a **local minimum** at its vertex.

Q2: Is this maximum also a global maximum?  
A2: Yes, for a downward-opening parabola ((a\<0)), the local maximum is also the **global maximum**.

Q3: Can we use completing the square instead?  
A3: Absolutely. Completing the square on (-x^2-2x+3) yields (-(x+1)^2+4), directly showing a maximum of **4** at (x=-1).

Would you like me to generate a few practice problems on finding the vertices of quadratics?  
@Gulce_Gedikli1