Respuesta :

we have

[tex]f(x)=-(x+3)(x+1)[/tex]

we know that

The function is a quadratic equation open downward

The vertex is a maximum

The zero's of the function are the points [tex](-3,0)[/tex] and[tex](-1,0)[/tex]

therefore

the answer in the attached figure

Ver imagen calculista

The graph of the function [tex]f\left( x \right)=- \left( {x + 3} \right)\left( {x + 1} \right)[/tex] is attached below. Graph (b) is correct.

Further explanation:

Given:

The function is [tex]f\left( x \right) =  - \left( {x + 3} \right)\left( {x + 1} \right).[/tex]

Explanation:

The given function is [tex]f\left( x \right) =  - \left( {x + 3} \right)\left( {x + 1} \right)[/tex].

The graph of the function represents downward parabola as the function has negative sign.

Solve the function to obtain the zeroes.

[tex]\begin{aligned}f\left( x \right)&= - \left( {x + 3} \right)\left( {x + 1} \right)\\0&= - \left( {x + 3} \right)\left( {x + 1} \right)\\0&= \left( {x + 3} \right){\text{or }}\left( {x + 1} \right) &= 0\\x &= - 3\;{\text{or}}\;x\;&= 0\\\end{aligned}[/tex]

The zeroes of the function are [tex]-3[/tex] and [tex]-1[/tex].

In graph 2 the zeros are [tex]-3[/tex] and [tex]-1[/tex].

The graph of the function [tex]f\left( x \right) =  - \left( {x + 3} \right)\left( {x + 1} \right)[/tex] is attached below. Graph (b) is correct.

Graph (a) doesn’t represent the graph of the function.

Graph (b) represents the graph of the function.

Graph (c) doesn’t represent the graph of the function.

Graph (d) doesn’t represent the graph of the function.

Learn more:

  1. Learn more about inverse of the function https://brainly.com/question/1632445.
  2. Learn more about equation of circle brainly.com/question/1506955.
  3. Learn more about range and domain of the function https://brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Polynomials

Keywords: quadratic equation, equation factorization. Factorized form, polynomial, quadratic formula, zeroes, Fundamental Theorem of algebra, polynomial.