Problem 1. Given a vertex and another point on a parabola, write in the form f (x) = a(x − h)2 + k. Find the zeros of this function by setting f(x) = 0 and solving for x. Rewrite each function in the form f(x) = ax2 + bx + c. Use the quadratic formula to find the zeros and the vertex of the parabola. Graphing the parabola may help you to keep all of this straight. It’s not necessary, but I highly recommend doing it so that you’re not just “plugging numbers” into the right places, you’re actually seeing why you’re plugging in inputs and getting outputs. An example is given below.
(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:2,2,Point:(1,1) 33
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
andwejustneedtofinda. Since(3,1)isapointonthefunction,y=1whenx=3. Thatis,the
f(x)=a(x−1)2 +2 output of f is 1 when the input is 3. Symbolically,
Using our f(x) from above, we have
f(3) = 1
f(3)=a(3−1)2 +2=1
⇐⇒ a(2)2 + 2 = 1 ⇐⇒ a4+2=1 ⇐⇒ 4a=−1
⇐ ⇒ a = − 41
Plugging in our value for a into our original f(x) gives us the answer to the first task:
f(x)=−41(x−1)2 +2 To find the zeros of this function we solve f(x) = 0 for x:
−14(x − 1)2 + 2 = 0 −14(x−1)2 =−2
(x − 1)2 = −2(−4)
(x−1)2 =8
(x − 1)2 = ±√8 √
x−1=± 2∗2∗2 √
x−1=±2 2 √
x=1±2 2
x ∈ 1 − 2√2, 1 + 2√2
Rewritingf(x)=−41(x−1)2 +2intheformf(x)=ax2 +bx+cgivesus f(x)=−41(x−1)2 +2
= −14(x2 − 2x + 1) + 2
= − 14 x 2 + 24 x − 14 + 2
= − 14 x 2 + 12 x − 14 + 48
= − 14 x 2 + 12 x + 74
Sof(x)=−14×2+12x+74 isinthedesiredformwherea=−14,b=21,andc=74.Pluggingthisin
√
tox=−b ± b2−4ac: 2a 2a
√ x=−b± b2−4ac
1
2a 2a
12 −4−17
244 = −2−1 ± 2−1
2
44
12 +47
244 = −−1 ± −1
22
1 2
= −(−1) ± − 1 2
1+(1)7 44
14 + 74 =1± −1
2
8
4 = 1 ± − 1
2
√
2 = 1 ± − 1
2
√ 2 =1± 2 −1
√ =1∓2 2
And therefore
Obviously the midpoint of these two zeros is 1 (which happens to be equal to −b . . . imagine that!)
x ∈ 1 − 2√2, 1 + 2√2
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 f(x) = −41×2 + 12x + 74 to get
f(1) = −14(1)2 + 12(1) + 74
= − 14 + 12 + 47 = − 14 + 24 + 47 = − 14 + 94
= 84 =2
Then the vertex is at (1, 2), as desired.
