The vertex form of a function is g(x) = (x - 3)2 +9. How does the graph of g(x) compare to the graph of the function
f(x) = x2?
g(x) is shifted 3 units left and 9 units up.
g(x) is shifted 3 units right and 9 units up.
g(x) is shifted 9 units left and 3 units down.
g(x) is shifted 9 units right and 3 units down.

the perimeter is increased by a factor of 2

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

hope this .

r3t40

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.