Section 3.2 Standard Form
Vertex Graphing Form: f(x) = a(x – p)2 + q
vertex is (p, q)
Standard Form: f(x) = ax2 + bx + c , a ≠ 0
If we expand the vertex graphing form, we can compare the 2 forms:
f(x) = a(x – p)2 + q
FOIL
f(x) = a(x2 – 2xp + p2) + q
2
2
f(x) = ax – 2axp + ap + q
OR
f(x) = ax2 + (-2ap)x + (ap2 + q) {Now place the standard form below and compare:
f(x) = ax2 + ( b )x + c
SO the a’s match up
b = -2ap and
c = ap2 + q
By rewriting b = -2ap, we can find a formula to find the “p” coordinate of the vertex:
b = -2ap ====>
p=−
b
2a
Example1: Find the vertex for the quadratic function f(x) = -x2 +2x + 8.
First identify a,b,c in the function: a= -1, b = 2, c = 8
b
If we find p, using the formula p = − , then we have half of the vertex (p, q).
2a
b
2
So p = −
=−
= 1 . The vertex is (1, ?)
2a
2(−1)
To find the second half of the vertex, substitute 1 for x in the original function.
f(x) = -x2 +2x + 8
f(-1) = -(1)2 + 2(1) + 8
= -1 + 2 + 8
=9
Therefore the vertex is at (1, 9)
Example2: Using the same function as in Example 1 answer the following:
a) Does the graph open UP or DOWN? Down, since “a” is negative.
b) Find the x and y intercepts.
y = -x2 +2x + 8 Put 0 in for x to find the y-int.
y = -(0)2 + 2(0) + 8
y = 8 {y-int is 8}
Hints when factoring:
Look for GCF -2x2 + 4x = -2x(x – 2)
y = -x2 +2x + 8 Put 0 in for y to find the x-int.
0 = -x2 + 2x + 8 {Now try to factor to solve}
0 = -1(x2 – 2x – 8)
0 = -1(x – 4)(x + 2)
x = 4 and x = -2 The two x-intercepts are 4 and -2.
Look for Diff of Squares
x2 – 4 = (x + 2)(x – 2)
Look for Trinomials x2 – 4x – 21 = (x – 7)(x + 3)
Example 3: A diver jumps from a 3-m springboard with an initial vertical velocity of 6.8m/s. Her
height, “h”, in metres, above the water “t” seconds after leaving the diving board can be
modeled by the function h(t) = -4.9t2 + 6.8t + 3.
a) What does the y-intercept represent?
The y-intercept represents the height after 0 seconds (put 0 in for t).
y = -4.9(0)2 + 6.8(0) + 3 = 3 metres This is the starting height, ie the height of the diving board.
b) What is the height of the diver 0.6 seconds after leaving the diving board?
Put 0.6 in for t to find the height h(0.6).
h(0.6) = -4.9(0.6)2 + 6.8(0.6) + 3
= 5.316
So the height of the diver after 0.6 sec is 5.3 metres.
c) What is the maximum height of the diver and when does she reach this height?
We need to find the vertex (p, q). “p” will be the time in seconds and “q” will be the height.
Use the formula p = −
p=-
6.8
= .6939
2(-4.9)
b
with a = -4.9 and b = 6.8
2a
So the vertex is (.6939, ?)
To find q, put .6939 in for “t” to find the height.
h(.6939) = -4.9(.6939)2 + 6.8(.6939) + 3
= 5.359
This means the vertex is (.6939, 5.359).
So our maximum height is at about 5.4 metres after about .7 seconds.
d) When does the diver hit the water?
The diver hits the water when the height is 0. This would be an x-intercept.
Put 0 in for the height and solve for t.
h(t) = -4.9t2 + 6.8t + 3
0 = -4.9t2 + 6.8t + 3 Use the quadratic formula for these types that are not factorable.
- 6.8 ± (6.8) 2 - 4(-4.9)(3)
t=
2(-4.9)
t=
quadratic formula x =
Careful here -4 times -4.9 = positive amount
- 6.8 ± 105.04
- 9.8
t = -.35 or 1.74
− b ± b 2 − 4ac
2a
You get 2 answers due to the ±. Only one makes sense in this case.
The diver hits the water after about 1.74 seconds.
e) State the domain and range for this problem.
Domain: {t | 0 ≤ t ≤ 1.74}
Range: {h | 0 ≤ h ≤ 5.4}
{Here we use the point where the diver hits the ground.}
{We need the vertex (.7, 5.4) for the range.}
Below is a graph of the above function:
h
7
6.5
6
5.5
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
t
Textbook Assignment: Page 174-176 #1all, 2all, 4all (graph using any method), 6all, 7, 11, 12