The function h(t) = −5t2 + 15t shown in the graph models the supporting structure of a bridge:
graph of parabola starting at zero comma zero, rising from the left to about one and one half comma 11, and falling to the right, ending at 3 comma zero
what is the domain of h(t)?
all real numbers
0 ≤ x ≤ 11
0 ≤ x ≤ 3
x ≥ 0

"All real numbers"

Step-by-step explanation:

The domain of a function is the set of x-values that the function is defined for. The set of x-values for which the function has y-values.

Allowed x-values, basically.

Looking at the graph, we would initially think the x values for which there is a function drawn would be from x= 0 to x = 3, but we must remember that this is a continuous function so it stretches upwards and downwards, both. So the set of x-values for which the function is defined is "set of all real numbers".

Normally, a function in the form f(x) = ax^2 + bx, where and b are real numbers, will be a parabola whose domain would be all real numbers.

The function h(t) is in such form, so the domain is "all real numbers".

step-by-step explanation:

from the graph attached below:

domain of graph is the value of x for which the given function profit function p(x) is defined.

(a) the domain represents the number of years for which the outlet will earn a profit.

range of profit function defines the lowest to highest value of profit earned.

(b) the range of this function is (0 , 160 ). for this function, the range represents the outlet's profit earned.

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!