A quadratic function y=f(x) is plotted on a graph and the vertex of the resulting parabola is (3,-4). What is the vertex of the function defined as g (x) =f(x+5)?

(-2, -4)

Step-by-step explanation:

y = f(x) = ax²+bx +c

given the vertex is (3, -4)

=> the symetry axis: x = 3

x = -b/2a = 3

so, -b = 6a => b = -6a

the function f(x) = x²-6x + 5

g(x) = f(x+5)

g(x) = (x+5)²-6(x+5)+5

g(x) = x²+10x+25-6x-30+5

g(x) = x²+4x

=> a=1, b=4

find the vertex of the function g :

the symetry axis: x= -4/2(1) => x = -2

y = g(-2) = (-2)²+4(-2)= 4-8 = -4

so, the vertex is (-2, -4)

your answer is the frist one

hop i