r
Section 7.2
1. The interesting points of a parabola are the
y-intercept, any x-intercepts, and the vertex.
3. a. The value of c appears on the graph as the
y-intercept, (0, c).
b. The value of b appears on the graph as the
y-intercept, (0, b).
c. The value of d appears on the graph as the
y-intercept, (0, d).
d. Substituting 0 or x into any of these equations
gives an output equal to the constant, and the
constant is the y-coordinate of the y-intercept.
@ Houghton Mifflin Company. All rights reserved.
-
--
----
~
--
~
_._-
Section 7.2
107
S. a. Yl = X"2- 5X+ 3, Y2 = 0
Windows may vary: Xmin = "9.4, Xmax = 9.4,
Ymin= "10,Ymax= 10
Because the x-intercepts occur where y
we may TRACE to where Y
:::: 0
=0
and note the
X-values, or find closer approximations by
entering Y2 = 0 in the Y= menu and using 2nd
CALC intersect followed by ENTER, ENTER,
ENTER"
There is no x-intercept because the parabola
does not cross the x-axis.
x-intercept
There is one x-intercept because the parabola
touches the x-axis once.
c. Yl =X"2-5
x-intercepts
Il'Ihr5~cti(JJJ
~='I.3027756
1'1'=0
Solutions: X::::.7 or X::::4.3
There are two x-intercepts because the
parabola crosses the x-axis twice.
9. a. y=-16x2+32x+48
y = -16(x2- 2x-3)
y= -16(x-3)(x+l)
b. y= -16(x-3)(x+l)
x-3=00rx+l-0
x = 3 orx = -1
c.
c. The value of Xmin determines the left edge of
the viewing screen and the value of Xmax the
right. To include the x-intercepts on a graph
of the parabola we need friendly values such
that Xmin < "I and Xmax > 3.
Windows may vary: Xmin = "4.7,Xmax= 4.7
Mil'lil'ilUM
~=2.500(l01
1'1'=-3,,:5
Turning point = (2.5, "3.25).
d. Minimum
d. The value ofYmin determines the bottom
edge of the viewing screen and Ymax the top.
We want our graph to include the x-axis and
the turning point. Increase Ymax until the
turning point appears.
Windows may vary: Ymin = -3.1,Ymax= 70
@ Houghton Mifflin Company. All rights reserved.
- -
-- ...
..
108
!!!
Chapter 7: Quadratic Equations and Graphs
11. a. The ball starts at 5.1 feet above ground, rises
to about 101.1 feet and drops again. The ball
can't have a negative height, so a value less
than 0 means the ball has landed. The ball hits
the ground between 5 and 6 seconds.
b. y = 2x2 + 7x-IS
0=2x2+7x-15
The expression on the right side of the
equation factors.
Diagonal product
b. The ball reaches its maximum height between
2 and 3 seconds.
13. a. Y1 = 16X"2 + 80X + 4
Windowsmayvary:Xmin= 0, Xmax= 9.4,
Ymin= 0, Ymax= 150
°
L\
M=2,SOOCII)U _'1'=10'1
No, the water does not reach 120 feet.
b. From the graph, the water will reach an
approximate height of 104 feet.
15. We are looking for the time it takes for the
football to land (h = 0). Thus we are trying to
solve 0 = "4.0r + 221+ 2. This quadratic doesn't
factor nicely so we can use a graph to
approximate the time,
Yl == °4.9X"2+ 22X...2
Windows may vary: Xmin = 0, Xmax = 9.4,
Ymin '" °S,Ymax · 30
= -30x2
-3x, lOx is the factor pair oC-30~ that adds to
7x.
o = 2x2 - 3x + lOx -IS
0= (2x2 - 3x) + (lOx -15)
0= x(2x-3) +5(2x -3)
0=(2x-3)(x+5)
2x-3=00rx+5=0
3
x=- orx=
2
_
5
19. x2 + 3x - 28 = 0
The expression on the left side of the equation can
be factored by splitting the middle term. The
diagonal product = °28x2.
"4x, 7x is the factor pair of "28x2that adds to 3x.
x2 -4x+7x-28
=0
Using an area diagram or factoring by grouping,
gives
(x-4)(x+7)=0
x-4=00rx+7=0
x = 4 or x = -7
:zI. (2y - 3)(2y...3)
..4y2+6y-6y-9
=4y2 - 9
23.
5xy.z
20x3 y2 Z
1-3 4-2 1-1
=x
Il'Ihru:c:ti(lfl
M=It.S7B93S3 ''1'=0
The footballwillremainairborneapproximately
4.6 seconds.
Y z
4
=X'2y2z0
4
-L
- 4x2
Skills and Review 7.2
25. a. The area of a rectangle is length
17. a. y=2x2+7x-15
=2C1)2+7C1)-15
= -20
>to
width
A = 8.5 inches >to11 inches
= 93.5 inches2
b. The perimeter is the distance around the
outside.
P = 8.5 in + II in + 8.5 in + 11 in
= 39 inches
@ Houghton Mifflin Company. All rights reserved.