Question 298717: I am having trouble with these three problems.
1. Without graphing, find the vertex.
f(x)= -(x + 3)^2 + 4
2. Without graphing, find the line of symmetry.
f(x)= 13/6(x-5)^2 + 5
3. Without graphing, find the minimum value.
f(x) = (x-1)^2 + 1
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website!
1. Without graphing, find the vertex.
f(x)= -(x + 3)^2 + 4
The "vertex form" of a quadratic/parabola is:
where
(h,k) is the vertex
.
So, starting with:
f(x)= -(x + 3)^2 + 4
we can rewrite to look like the "vertex form":
f(x)= -(x-(-3))^2 + 4
From inspection the
vertex is at (-3,4)
.
2. Without graphing, find the line of symmetry.
f(x)= 13/6(x-5)^2 + 5
Since the above is already in "vertex form" we immediately see that
vertex is at (-5, 5)
"line is symmetry" is then
x = -5
.
3. Without graphing, find the minimum value.
f(x) = (x-1)^2 + 1
Since the above is already in "vertex form" we immediately see that
vertex is at (-5, 5)
Since the coefficient associated withe the x^2 term is positive, we know it is a parabola that opens upwards. The vertex is then the minimum.
Therefore,
minimum value is k (of the (h,k) vertex)
Minimum is 5
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