Answer:
[tex]f"'(x) = 2(x-4)^2 - 1[/tex]
Step-by-step explanation:
Given
[tex]f(x) = x^2[/tex] --- the quadratic function
Vertically stretched by 2
Translation: [tex]4\ units[/tex] left and [tex]1\ unit[/tex] down
Required
Determine the new function
[tex]f(x) = x^2[/tex]
The rule for vertical stretch is:
[tex](x,y) \to (x,ay)[/tex]
In this case:
[tex]a = 2[/tex]
So, we have:
[tex]f(x) = x^2[/tex]
[tex]f'(x) = a * f(x)[/tex]
[tex]f'(x) = a * x^2[/tex]
Substitute: [tex]a = 2[/tex]
[tex]f'(x) = 2 * x^2[/tex]
[tex]f'(x) =2x^2[/tex]
Translation: 4 units left
The rule is:
[tex](x,y) \to (x - c,y)[/tex]
In this case: [tex]c =4[/tex]
So, we have:
[tex]f"(x) = 2(x - 4)^2[/tex]
Translation: 1 unit down
The rule is:
[tex](x,y) \to (x,y-d)[/tex]
In this case, [tex]d=1[/tex]
So, we have:
[tex]f"'(x) = f"(x) - 1[/tex]
[tex]f"'(x) = 2(x-4)^2 - 1[/tex]
