Me pueden ayudar con esto porfa, es urgente 1)Defina gráficamente una función en el intervalo [-2, 1], continua en todo punto excepto en x = 0, positiva en x = 1 y negativa en x = -2 y que no tenga raíces.

2) Supongamos que f(x) es continua en el intervalo [-1, 1] y que f (-1) = 2, f (0) = -1 y f (1) = 3. ¿Cuál es el mínimo número de raíces que f(x) puede tener en ese intervalo? Interprete gráficamente

what is all that apply?

income. 10x+1300

expenses 3x+(1800+1500)

expenses 3x+3300

to break even:

income = expense,

10x + 1300 = 3x + 3300,

solve for x.

step-by-step explanation:

mark brainliest

the height of the surface increases from the center out to the sides of the road.

step-by-step explanation:

as we move out we move away from the x-axis.

and as mentioned in the problem statement that height of a road increases when we move away from the center x.

hence proved that 3rd option

the height of the surface increases from the center out to the sides of the road.

is the correct option.

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.