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.
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