Question

Which polynomial function could be represented by the graph below?
PLEASE ANSWER FAST IM TIMED
a. f(x) = x2 + x – 2
b. f(x) = 2x2 + 2x – 4
c. f(x) = x2 – x – 2
d. f(x) = 2x2 –2x – 4

Answer 1

The parabola passes through the points (-2,0), (0,-4), and (1,0), then these points must satisfy the equation of the parabola.

For example:
1) (0,-4)→for x=0, y=-4→The independent term must be equal to -4
Only option b and d have independent term of -4

2) (1,0)→for x=1, y=0
Equation b. f(1)=2(1)^2+2(1)-4=2(1)+2-4=2+2-4→f(1)=0  Ok
Equation d. f(1)=2(1)^2-2(1)-4=2(1)-2-4=2-2-4→f(1)=-4   No

Answer: Option b. f(x)=2x^2+2x-4 

Answer 2

The polynomial function is $\boxed{f\left( x \right) = 2{x^2} + 2x - 2}$ that is represented by the graph.

Further explanation:

Given:

The options of the equations are as follows.

1.$f\left( x \right) = {x^2} + x - 2$

2.$f\left( x \right) = 2{x^2} + 2x - 4$

3.$f\left( x \right) = {x^2} - x - 2$

4.$f\left( x \right) = 2{x^2} - 2x - 4$

Explanation:

The graph passes through the points $\left( {0, - 4} \right)$ and $\left( { - 2,0} \right).$

Substitute 0 for x and -4 for $f\left( x \right)$ in equation $f\left( x \right) = {x^2} + x - 2$ to check the point satisfy the equation.

$\begin{aligned}- 4&={\left( 0 \right)^2}+ 0 - 2\\- 4&\ne- 2\\\end{aligned}$

The point doesn’t satisfy the equation.

Substitute 0 for x and -4 for $f\left( x \right)$ in equation $f\left( x \right) = 2{x^2} + 2x - 4$ to check the point satisfy the equation.

$\begin{aligned}- 4&= 2{\left( 0 \right)^2} + 2\left( 0 \right)- 4\\- 4&=- 4\\\end{aligned}$

The point satisfies the equation.

Substitute -2 for x and 0 for $f\left( x \right)$ in equation $f\left( x \right) = 2{x^2} + 2x -4$ to check the point satisfy the equation.

$\begin{aligned}0&=2{\left( { - 2}\right)^2} + 2\left( { - 2} \right) - 4\\0&= 8 - 4 - 4\\0&= 0\\\end{aligned}$

The point satisfies the equation.

Substitute 0 for x and -4 for $f\left( x \right)$ in equation  to check $f\left( x \right) = {x^2} - x - 2$ the point satisfy the equation.

$\begin{aligned}- 4&= {\left( 0 \right)^2}- \left( 0 \right) - 2\\- 4&\ne- 2\\\end{aligned}$

The point doesn’t satisfy the equation.

Substitute 0 for x and -4 for $f\left( x \right)$ in equation $f\left( x \right) = 2{x^2} - 2x - 4$ to check the point satisfy the equation.

$\begin{aligned}- 4&= 2{\left(0 \right)^2} - 2\left( 0 \right) - 4\\- 4&=- 4\\\end{aligned}$

The point satisfies the equation.

Substitute -2 for x and 0 for $f\left( x \right)$ in equation $f\left( x \right) = 2{x^2} - 2x – 4$ to check the point satisfy the equation.

$\begin{galigned}0&= 2{\left( { - 2} \right)^2} - 2\left( { - 2} \right) - 4\\0&= 8 + 4 - 4\\0\ne8\\\end{aligned}$

The point doesn’t satisfy the equation.

Hence, the polynomial function is $\boxed{f\left( x \right) = 2{x^2} + 2x - 4}$ that is represented by the graph.Option (b) is correct.

Option (a) is not correct.

Option (b) is correct.

Option (c) is not correct.

Option (d) is not correct.

Learn more:

1. Learn more about inverse of the functionhttps://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, polynomial, quadratic formula, zeroes, function.