On a coordinate plane, a curved line with minimum values of (negative 1.56, negative 6) and (3, 0), and a maximum value of (1.2, 2.9), crosses the x-axis at (negative 2.5, 0), (0, 0), and (3, 0), and crosses the y-axis at (0, 0). Which interval for the graphed function has a local minimum of 0? [–3, –2] [–2, 0] [1, 2] [2, 4]

Step-by-step explanation:

On a coordinate plane, a curved line has minimum values of (negative 1.56, negative 6) and (3, 0). This means

1) when , the minimum value of the function is

2) when the minimum value of the function is

Hence, the graphed function has a local minimum of 0, when

Therefore, the interval which contains this value of x is because

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.

cross multiplying, we get :

⇒ (3) × (5) = (9) × (x)

⇒ 15 = 9x