Write A Quadratic Function With Zeros -9 And -4.Write Your Answer Using The Variable { X$}$ And In Standard Form.${ F(x) = }$ { \square \} ${ 2 \square 4 }$

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Understanding Quadratic Functions

Quadratic functions are a type of polynomial function that can be written in the form of f(x)=ax2+bx+c{ f(x) = ax^2 + bx + c }, where a{ a }, b{ b }, and c{ c } are constants, and a≠0{ a \neq 0 }. The zeros of a quadratic function are the values of x{ x } that make the function equal to zero. In this article, we will learn how to write a quadratic function with zeros -9 and -4.

The Factorized Form of a Quadratic Function

A quadratic function can be written in the factorized form as f(x)=a(xβˆ’r1)(xβˆ’r2){ f(x) = a(x - r_1)(x - r_2) }, where r1{ r_1 } and r2{ r_2 } are the zeros of the function. Given that the zeros are -9 and -4, we can write the function as f(x)=a(x+9)(x+4){ f(x) = a(x + 9)(x + 4) }.

The Standard Form of a Quadratic Function

To write the function in standard form, we need to expand the factorized form. Using the distributive property, we can expand the function as follows:

f(x)=a(x+9)(x+4){ f(x) = a(x + 9)(x + 4) } f(x)=a(x2+4x+9x+36){ f(x) = a(x^2 + 4x + 9x + 36) } f(x)=a(x2+13x+36){ f(x) = a(x^2 + 13x + 36) }

Choosing the Value of a

Since the coefficient of the quadratic term is a{ a }, we can choose any value for a{ a } as long as it is not equal to zero. For simplicity, let's choose a=2{ a = 2 }. This gives us the quadratic function:

f(x)=2(x2+13x+36){ f(x) = 2(x^2 + 13x + 36) }

Simplifying the Function

To simplify the function, we can distribute the coefficient a{ a } to the terms inside the parentheses:

f(x)=2x2+26x+72{ f(x) = 2x^2 + 26x + 72 }

Conclusion

In this article, we learned how to write a quadratic function with zeros -9 and -4. We started by writing the function in the factorized form, and then expanded it to the standard form. We also chose a value for the coefficient a{ a } and simplified the function. The final answer is:

f(x)=2x2+26x+72{ f(x) = 2x^2 + 26x + 72 }

Example Use Case

Suppose we want to find the value of the function at x=βˆ’5{ x = -5 }. We can plug in the value of x{ x } into the function:

f(βˆ’5)=2(βˆ’5)2+26(βˆ’5)+72{ f(-5) = 2(-5)^2 + 26(-5) + 72 } f(βˆ’5)=2(25)βˆ’130+72{ f(-5) = 2(25) - 130 + 72 } f(βˆ’5)=50βˆ’130+72{ f(-5) = 50 - 130 + 72 } f(βˆ’5)=βˆ’8{ f(-5) = -8 }

Therefore, the value of the function at x=βˆ’5{ x = -5 } is -8.

Tips and Variations

  • To find the zeros of a quadratic function, we can set the function equal to zero and solve for x{ x }.
  • To find the maximum or minimum value of a quadratic function, we can use the vertex formula: x=βˆ’b2a{ x = -\frac{b}{2a} }.
  • To graph a quadratic function, we can use the x-intercepts and the vertex to draw the graph.

Common Mistakes

  • Not choosing a value for the coefficient a{ a }.
  • Not expanding the factorized form correctly.
  • Not simplifying the function correctly.

Conclusion

In conclusion, writing a quadratic function with zeros -9 and -4 requires us to write the function in the factorized form, expand it to the standard form, and choose a value for the coefficient a{ a }. We also learned how to simplify the function and find the value of the function at a given point. The final answer is:

f(x)=2x2+26x+72{ f(x) = 2x^2 + 26x + 72 }

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

A: The factorized form of a quadratic function is f(x)=a(xβˆ’r1)(xβˆ’r2){ f(x) = a(x - r_1)(x - r_2) }, where r1{ r_1 } and r2{ r_2 } are the zeros of the function.

Q: How do I find the zeros of a quadratic function?

A: To find the zeros of a quadratic function, you can set the function equal to zero and solve for x{ x }. For example, if we have the function f(x)=a(x+9)(x+4){ f(x) = a(x + 9)(x + 4) }, we can set it equal to zero and solve for x{ x }:

a(x+9)(x+4)=0{ a(x + 9)(x + 4) = 0 }

Q: How do I expand the factorized form of a quadratic function?

A: To expand the factorized form of a quadratic function, you can use the distributive property. For example, if we have the function f(x)=a(x+9)(x+4){ f(x) = a(x + 9)(x + 4) }, we can expand it as follows:

f(x)=a(x+9)(x+4){ f(x) = a(x + 9)(x + 4) } f(x)=a(x2+4x+9x+36){ f(x) = a(x^2 + 4x + 9x + 36) } f(x)=a(x2+13x+36){ f(x) = a(x^2 + 13x + 36) }

Q: How do I simplify the function?

A: To simplify the function, you can distribute the coefficient a{ a } to the terms inside the parentheses. For example, if we have the function f(x)=a(x2+13x+36){ f(x) = a(x^2 + 13x + 36) }, we can simplify it as follows:

f(x)=a(x2+13x+36){ f(x) = a(x^2 + 13x + 36) } f(x)=ax2+13ax+36a{ f(x) = ax^2 + 13ax + 36a }

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

A: The standard form of a quadratic function is f(x)=ax2+bx+c{ f(x) = ax^2 + bx + c }, where a{ a }, b{ b }, and c{ c } are constants, and a≠0{ a \neq 0 }.

Q: How do I find the value of a quadratic function at a given point?

A: To find the value of a quadratic function at a given point, you can plug in the value of x{ x } into the function. For example, if we have the function f(x)=2x2+26x+72{ f(x) = 2x^2 + 26x + 72 } and we want to find the value at x=βˆ’5{ x = -5 }, we can plug in the value of x{ x } as follows:

f(βˆ’5)=2(βˆ’5)2+26(βˆ’5)+72{ f(-5) = 2(-5)^2 + 26(-5) + 72 } f(βˆ’5)=2(25)βˆ’130+72{ f(-5) = 2(25) - 130 + 72 } f(βˆ’5)=50βˆ’130+72{ f(-5) = 50 - 130 + 72 } f(βˆ’5)=βˆ’8{ f(-5) = -8 }

Q: What is the vertex formula for a quadratic function?

A: The vertex formula for a quadratic function is x=βˆ’b2a{ x = -\frac{b}{2a} }, where a{ a } and b{ b } are the coefficients of the quadratic function.

Q: How do I graph a quadratic function?

A: To graph a quadratic function, you can use the x-intercepts and the vertex to draw the graph. The x-intercepts are the points where the function crosses the x-axis, and the vertex is the point where the function reaches its maximum or minimum value.

Q: What are some common mistakes to avoid when working with quadratic functions?

A: Some common mistakes to avoid when working with quadratic functions include:

  • Not choosing a value for the coefficient a{ a }.
  • Not expanding the factorized form correctly.
  • Not simplifying the function correctly.

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

A: To find the maximum or minimum value of a quadratic function, you can use the vertex formula: x=βˆ’b2a{ x = -\frac{b}{2a} }. This will give you the x-coordinate of the vertex, and you can then plug this value into the function to find the corresponding y-coordinate.

Q: Can I use a calculator to find the zeros of a quadratic function?

A: Yes, you can use a calculator to find the zeros of a quadratic function. Simply enter the function into the calculator and use the zero-finding feature to find the zeros.

Q: How do I check my work when working with quadratic functions?

A: To check your work when working with quadratic functions, you can plug in the values of x{ x } into the function and verify that the results are correct. You can also use a calculator to check your work.

Q: What are some real-world applications of quadratic functions?

A: Quadratic functions have many real-world applications, including:

  • Modeling the trajectory of a projectile
  • Finding the maximum or minimum value of a function
  • Solving problems involving optimization
  • Modeling the growth or decay of a population

Q: Can I use quadratic functions to model real-world data?

A: Yes, you can use quadratic functions to model real-world data. Quadratic functions can be used to model a wide range of phenomena, including the growth or decay of a population, the trajectory of a projectile, and the maximum or minimum value of a function.

Q: How do I choose the correct quadratic function to model real-world data?

A: To choose the correct quadratic function to model real-world data, you need to consider the characteristics of the data and the type of function that best fits the data. You can use statistical methods, such as regression analysis, to determine the best-fitting quadratic function.