Given is the function f(x) = -x2

a. The graph of f shifted 3 units down. The graph that comes into existence, belongs to the function g. Calculate the points of intersection of the graph of g with the line y = -28

b. The graph of h starts from the graph of f and has been shifted up a few unit. Then the graph is multiplied with the factor 2. The function notation for the new graph is h(x) = -2x2 + 6. Calculate how many units graph of f was shifted up to form the new graph.

Guest Jun 10, 2019

#1**+1 **

Given is the function f(x) = -x2

a. The graph of f shifted 3 units down. The graph that comes into existence, belongs to the function g. Calculate the points of intersection of the graph of g with the line y = -28

The new function is

g(x) = -x^2 - 3 set this = to 28

-x^2 - 3 = -28 add 3 to both sides

-x^2 = - 25 multiply through by - 1

x^2 = 25 take both roots

x = ±√25

x = ±5

So....the points of intersection are (5, -28) and (-5, 28)

b. The graph of h starts from the graph of f and has been shifted up a few unit. Then the graph is multiplied with the factor 2. The function notation for the new graph is h(x) = -2x2 + 6. Calculate how many units graph of f was shifted up to form the new graph.

Let the sfift of f be c units

So we have that f(x) - -x^2 + c

2f(x) = 2 [-x^2 + c ] = -2x^2 + 6

-2x^2 + 2c = - 2x^2 + 6

So

2c = 6

c = 3 units

CPhill Jun 10, 2019