Description

(a) Vertex: (−2,3), Point: (−1,2)
(b) Vertex: (3,−1), Point: (0,1)
(c) Vertex: (2,1), Point: (1,2)
(d) Vertex: (−5,−2), Point: (−1,6)
(e) Vertex: , Point: (1,1)
Example. Vertex: (1,2), Point: (3,1)
Since the vertex is at (1,2), we can realize this function as the graph of x2 shifted right one unit and up 2 units. Therefore f(x) = a(x −1)2+2
and we just need to find a. Since (3,1) is a point on the function, y = 1 when x = 3. That is, the output of f is 1 when the input is 3. Symbolically,
f(3) = 1
Using our f(x) from above, we have
f(3) = a(3−1)2+2 = 1
⇐⇒ a(2)2+2 = 1

Plugging in our value for a into our original f(x) gives us the answer to the first task:

To find the zeros of this function we solve f(x) = 0 for x:

Rewriting in the form f(x) = ax2+ bx + c gives us

So is in the desired form where , and . Plugging this in to :

And therefore
Obviously the midpoint of these two zeros is 1 (which happens to be equal to imagine that!) and therefore the x-value of the vertex is x = 1. To find the vertex’s y-value, we plug in x = 1 into our function to get

Then the vertex is at (1,2), as desired.