The Function F ( X ) = − ( X − 3 ) 2 + 9 F(x)=-(x-3)^2+9 F ( X ) = − ( X − 3 ) 2 + 9 Can Be Used To Represent The Area Of A Rectangle With A Perimeter Of 12 Units, As A Function Of The Length Of The Rectangle, X X X . What Is The Maximum Area Of The Rectangle?A. 3 Square Units B. 6 Square

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Introduction

In mathematics, functions are used to represent various real-world phenomena. One such function is the area of a rectangle with a given perimeter. In this article, we will explore the function f(x)=−(x−3)2+9f(x)=-(x-3)^2+9 and determine its maximum value, which represents the maximum area of the rectangle.

Understanding the Function

The given function is f(x)=−(x−3)2+9f(x)=-(x-3)^2+9. This function represents the area of a rectangle with a perimeter of 12 units, as a function of the length of the rectangle, xx. To understand this function, let's break it down into its components.

The function f(x)f(x) is a quadratic function, which means it has a parabolic shape. The general form of a quadratic function is f(x)=ax2+bx+cf(x)=ax^2+bx+c. In this case, the function is f(x)=−(x−3)2+9f(x)=-(x-3)^2+9, which can be rewritten as f(x)=−x2+6x−9+9f(x)=-x^2+6x-9+9. Simplifying this expression, we get f(x)=−x2+6xf(x)=-x^2+6x.

The Meaning of the Function

The function f(x)f(x) represents the area of a rectangle with a perimeter of 12 units. The perimeter of a rectangle is given by the formula P=2l+2wP=2l+2w, where ll is the length and ww is the width. Since the perimeter is fixed at 12 units, we can write the equation 2l+2w=122l+2w=12. Solving for ww, we get w=6−lw=6-l.

Substituting the Expression for Width

Now that we have an expression for the width, we can substitute it into the function f(x)=−x2+6xf(x)=-x^2+6x. This gives us f(x)=−x2+6x+6−xf(x)=-x^2+6x+6-x. Simplifying this expression, we get f(x)=−x2+6x+6−xf(x)=-x^2+6x+6-x.

Simplifying the Function

Simplifying the function further, we get f(x)=−x2+5x+6f(x)=-x^2+5x+6.

Finding the Maximum Value

To find the maximum value of the function, we need to find the vertex of the parabola. The vertex of a parabola is given by the formula x=−b2ax=-\frac{b}{2a}. In this case, a=−1a=-1 and b=5b=5. Plugging these values into the formula, we get x=−52(−1)x=-\frac{5}{2(-1)}. Simplifying this expression, we get x=52x=\frac{5}{2}.

Substituting the Value of x

Now that we have the value of xx, we can substitute it into the function f(x)=−x2+5x+6f(x)=-x^2+5x+6. This gives us f(x)=−(52)2+5(52)+6f(x)=-\left(\frac{5}{2}\right)^2+5\left(\frac{5}{2}\right)+6. Simplifying this expression, we get f(x)=−254+252+6f(x)=-\frac{25}{4}+\frac{25}{2}+6.

Evaluating the Expression

Evaluating the expression further, we get f(x)=−254+504+244f(x)=-\frac{25}{4}+\frac{50}{4}+\frac{24}{4}. Simplifying this expression, we get f(x)=494f(x)=\frac{49}{4}.

Converting the Fraction to a Decimal

Converting the fraction to a decimal, we get f(x)=12.25f(x)=12.25.

Conclusion

In conclusion, the maximum area of the rectangle is 12.25 square units.

Final Answer

The final answer is 12.25.

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Introduction

In our previous article, we explored the function f(x)=−(x−3)2+9f(x)=-(x-3)^2+9 and determined its maximum value, which represents the maximum area of the rectangle. In this article, we will answer some frequently asked questions related to this topic.

Q: What is the function f(x)=−(x−3)2+9f(x)=-(x-3)^2+9 used for?

A: The function f(x)=−(x−3)2+9f(x)=-(x-3)^2+9 is used to represent the area of a rectangle with a perimeter of 12 units, as a function of the length of the rectangle, xx.

Q: How do you find the maximum value of the function?

A: To find the maximum value of the function, we need to find the vertex of the parabola. The vertex of a parabola is given by the formula x=−b2ax=-\frac{b}{2a}. In this case, a=−1a=-1 and b=5b=5. Plugging these values into the formula, we get x=−52(−1)x=-\frac{5}{2(-1)}. Simplifying this expression, we get x=52x=\frac{5}{2}.

Q: What is the maximum area of the rectangle?

A: The maximum area of the rectangle is 12.25 square units.

Q: How do you convert the fraction to a decimal?

A: To convert the fraction to a decimal, we can divide the numerator by the denominator. In this case, we get f(x)=494=12.25f(x)=\frac{49}{4}=12.25.

Q: What is the significance of the function f(x)=−(x−3)2+9f(x)=-(x-3)^2+9?

A: The function f(x)=−(x−3)2+9f(x)=-(x-3)^2+9 is significant because it represents the area of a rectangle with a perimeter of 12 units, as a function of the length of the rectangle, xx. This function can be used to model real-world problems involving rectangles and their areas.

Q: How do you simplify the function f(x)=−(x−3)2+9f(x)=-(x-3)^2+9?

A: To simplify the function, we can start by expanding the squared term. This gives us f(x)=−x2+6x−9+9f(x)=-x^2+6x-9+9. Simplifying this expression, we get f(x)=−x2+6xf(x)=-x^2+6x.

Q: What is the general form of a quadratic function?

A: The general form of a quadratic function is f(x)=ax2+bx+cf(x)=ax^2+bx+c.

Q: How do you find the vertex of a parabola?

A: To find the vertex of a parabola, we can use the formula x=−b2ax=-\frac{b}{2a}. This formula gives us the x-coordinate of the vertex.

Q: What is the significance of the x-coordinate of the vertex?

A: The x-coordinate of the vertex represents the value of xx at which the function has its maximum or minimum value.

Q: How do you find the maximum or minimum value of a function?

A: To find the maximum or minimum value of a function, we can substitute the x-coordinate of the vertex into the function and evaluate it.

Q: What is the maximum or minimum value of the function f(x)=−(x−3)2+9f(x)=-(x-3)^2+9?

A: The maximum value of the function f(x)=−(x−3)2+9f(x)=-(x-3)^2+9 is 12.25 square units.

Conclusion

In conclusion, the function f(x)=−(x−3)2+9f(x)=-(x-3)^2+9 is used to represent the area of a rectangle with a perimeter of 12 units, as a function of the length of the rectangle, xx. The maximum area of the rectangle is 12.25 square units.

Final Answer

The final answer is 12.25.