Question
Find the vertex and axis of symmetry of the graph of a function.
f(x)=x^2+6x+16
f(x)=x^2+6x+16
Answers
drwls
f(x) = x^2+6x+16
= x^2 +6x +9 +3
= (x+3)^2 +3
The vertex (lowest point) is x= -3, y = 3. That is because (x+3)^2 can never ne negative.
It is symmetrical about the x=-3 vertical line.
= x^2 +6x +9 +3
= (x+3)^2 +3
The vertex (lowest point) is x= -3, y = 3. That is because (x+3)^2 can never ne negative.
It is symmetrical about the x=-3 vertical line.
Henry
f(x) = x^2 + 6x + 16.
h = Xv = -b / 2a = -6 / 2 = -3,
Substitute -3 for x in the given Eq:
k = Yv = (-3)^2 + 6*-3 16,
k = 9 - 18 + 16 = 25 - 18 = 7,
V(h , k) = V(-3 , 7).
Axis: h = Xv = -3. = A vertical line
where x isconstant at -3 for all values of y.
h = Xv = -b / 2a = -6 / 2 = -3,
Substitute -3 for x in the given Eq:
k = Yv = (-3)^2 + 6*-3 16,
k = 9 - 18 + 16 = 25 - 18 = 7,
V(h , k) = V(-3 , 7).
Axis: h = Xv = -3. = A vertical line
where x isconstant at -3 for all values of y.