Find the absolute maximum and absolute minimum values of f on the given interval. (If an answer does not exist, enter DNE.)
f(x) =
−490x/ x^2 + 49

on
[−11, 0].

x=-2

infinite number of lines, so there is an infinite number of equations.

step-by-step explanation:

there are an infinite number of line that passes through the point (-2,5)

x=-2   is one line that passes through the point

y = 5 is another line that passes through the point

(these two lines are perpendicular to each other).

since the number of   numbers are infinite and the slope is the ratio of two numbers,(change in y over change in x) we can have an infinite number of slopes.     our only condition is that it passes through the point (-2,5)

i think is 3 and 6. this took me 10 min to think.hope to and hope that my answer is brainliest.(follow me )

answer: evaluate the objective function at each vertex to determine which - and -values, if any, maximize or minimize the function. example: find the minimum value and maximum value of the objective function f ( x , y ) = 4 x + 5 y , subject to the following constraints

(- 1, 0 )

step-by-step explanation:

an x- intercept is where the function crosses the x- axis.

to find the x- intercept/s look for the value of x when f(x) = 0

from the given table there are 3 values of x where f(x) = 0

when x = - 1, x = 2, x = 3

the one that is listed is (- 1, 0 )