Note that
1.
[tex]f(x)=x^3+x^2-6x=x(x^2+x-6)=x(x-2)(x+3)[/tex]
The x-intercepts are at points x=-3, x=0, x=2. The graph should be increasing - decreasing - increasing.
2.
[tex]f(x)=x^3-x^2-6x=x(x^2-x-6)=x(x+2)(x-3)[/tex]
The x-intercepts are at points x=-2, x=0, x=3. The graph should be increasing - decreasing - increasing.
3.
[tex]f(x)=-2x^3-2x^2+12x=-2x(x^2+x-6)=-2x(x-2)(x+3)[/tex]
The x-intercepts are at points x=-3, x=0, x=2. The graph should be decreasing - increasing - decreasing.
4.
[tex]f(x)=-2x^3+x^2+12x=-2x(x^2-x-6)=-2x(x+2)(x-3)[/tex]
The x-intercepts are at points x=-2, x=0, x=3. The graph should be decreasing - increasing - decreasing.
From the diagram you can see that x-intercepts are at points x=-3, x=0, x=2 and the graph is decreasing-increasing-decreasing.
Answer: correct choice is 3.
The polynomial function is [tex]\boxed{f\left( x \right) = - 2{x^3} - 2{x^2} + 12x}[/tex] that is represented by the graph. Option (3) is correct.
Further explanation:
Given:
The options of the equations are as follows.
1.[tex]f\left( x \right) = {x^3} + {x^2} - 6x[/tex]
2. [tex]f\left( x \right) = {x^3} - {x^2} - 6x[/tex]
3. [tex]f\left( x \right) = - 2{x^3} - 2{x^2} + 12x[/tex]
4. [tex]f\left( x \right) = - 2{x^3} + 2{x^2} + 12x[/tex]
Explanation:
The graph passes through the points [tex]\left( {-3, 0}\right)[/tex] and [tex]\left( { 2,0} \right).[/tex]
Solve the polynomial [tex]f\left( x \right) = {x^3} + {x^2} - 6x[/tex] to obtain the zeros of x.
[tex]\begin{aligned}f\left( x \right)&= {x^3} + {x^2} - 6x\\&= x\left({{x^2} + x - 6}\right)\\&= x\left({x - 2}\right)\left({x + 3}\right)\\\end{aligned}[/tex]
The zeros of the polynomial are -3, 0 and 2.
The graph of the polynomial [tex]f\left( x \right) = {x^3} + {x^2} - 6x[/tex] is increasing-decreasing-increasing.
Solve the polynomial [tex]f\left( x \right) = {x^3} - {x^2} - 6x[/tex] to obtain the zeros of x.
[tex]\begin{aligned}f\left( x \right)&={x^3} - {x^2} - 6x\\&= x\left({{x^2} - x - 6}\right)\\&= x\left({x + 2} \right)\left({x - 3} \right)\\\end{aligned}[/tex]
The zeros of the polynomial are -2, 0 and 3.
The graph of the polynomial [tex]f\left( x \right) = {x^3} - {x^2} - 6x[/tex] is increasing-decreasing-increasing.
The graph doesn’t passes through the point [tex]\left({ - 3,0} \right).[/tex]Therefore, the polynomial doesn’t satisfy the graph.
Solve the polynomial [tex]f\left( x \right)= - 2{x^3} - 2{x^2} + 12x[/tex] to obtain the zeros of x.
[tex]\begin{aligned}f\left( x \right)&= - 2{x^3} + {x^2} + 12x\\&= - 2x\left({{x^2} + x - 6} \right)\\&= - 2x\left( {x - 2}\right)\left({x + 3}\right)\\\end{aligned}[/tex]
The zeros of the polynomial are -2, 0 and 3.
The graph of the polynomial [tex]f\left( x \right)=- 2{x^3} - 2{x^2} + 12x[/tex] is decreasing-increasing-decreasing.
Solve the polynomial [tex]f\left( x \right)= - 2{x^3} + 2{x^2} + 12x[/tex] to obtain the zeros of x.
[tex]\begin{aligned}f\left( x \right)&= - 2{x^3} + 2{x^2} + 12x\\&=- 2x\left( {{x^2} - x - 6} \right)\\&=- 2x\left({x + 2} \right)\left({x - 3} \right)\\\end{aligned}[/tex]
The zeros of the polynomial are -2, 0 and 3.
The graph of the polynomial [tex]f\left( x \right) = - 2{x^3} + 2{x^2} + 12x[/tex] is decreasing-increasing-decreasing.
The graph doesn’t passes through the point [tex]\left({ - 3,0}\right).[/tex] Therefore, the polynomial doesn’t satisfy the graph.
From the graph it has been observed that the graph is decreasing-increasing-decreasing.
The polynomial function is [tex]\boxed{f\left( x \right)= - 2{x^3} - 2{x^2} +12x}[/tex] that is represented by the graph. Option (3) is correct.
Learn more:
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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, polynomial, quadratic formula, zeroes, function.
