Graph the solution to each set of inequalities.
-
{
y
<
−
3
x
2
+
1
y
>
x
2
−
x
−
5
left brace . table with 2 rows and 1 column , row1 column 1 , y less than negative 3 , x squared , plus 1 , row2 column 1 , y greater than , x squared , minus x minus 5 , end table
-
{
y
<
3
x
2
+
2
x
+
1
y
>
2
x
2
−
x
+
1
left brace . table with 2 rows and 1 column , row1 column 1 , y less than 3 , x squared , plus 2 x plus 1 , row2 column 1 , y greater than 2 , x squared , minus x plus 1 , end table
-
{
y
>
x
2
−
5
x
+
4
y
>
x
2
+
3
x
+
2
left brace . table with 2 rows and 1 column , row1 column 1 , y greater than , x squared , minus 5 x plus 4 , row2 column 1 , y greater than , x squared , plus 3 x plus 2 , end table
-
Error Analysis A classmate graphed the system of inequalities and concluded that because the shaded regions do not intersect, there are no solutions to the system. Describe and correct the error.
{
y
≤
x
2
−
4
x
+
6
y
≥
x
2
−
4
x
+
2
left brace . table with 2 rows and 1 column , row1 column 1 , y less than or equal to , x squared , minus 4 x plus 6 , row2 column 1 , y greater than or equal to , x squared , minus 4 x plus 2 , end table
Image Long Description
Solve each system.
-
{
y
=
3
x
2
−
2
x
−
1
y
=
x
−
1
left brace . table with 2 rows and 1 column , row1 column 1 , y equals 3 , x squared , minus 2 x minus 1 , row2 column 1 , y equals x minus 1 , end table
-
{
y
=
−
x
2
+
x
−
5
y
=
x
−
5
left brace . table with 2 rows and 1 column , row1 column 1 , y equals negative , x squared , plus x minus 5 , row2 column 1 , y equals x minus 5 , end table
-
{
y
=
x
2
−
3
x
−
2
y
=
4
x
+
28
left brace . table with 2 rows and 1 column , row1 column 1 , y equals , x squared , minus 3 x minus 2 , row2 column 1 , y equals 4 x plus 28 , end table
-
{
y
=
1
2
x
2
+
4
x
+
4
y
=
−
4
x
+
12
1
2
left brace . table with 2 rows and 1 column , row1 column 1 , y equals , 1 half , x squared , plus 4 x plus 4 , row2 column 1 , y equals negative 4 x plus 12 , and 1 half , end table
-
{
y
=
−
3
4
x
2
−
4
x
y
=
3
x
+
8
left brace . table with 2 rows and 1 column , row1 column 1 , y equals negative , 3 fourths , x squared , minus 4 x , row2 column 1 , y equals 3 x plus 8 , end table
-
{
y
=
−
1
4
x
2
+
x
+
1
y
=
x
−
4
left brace . table with 2 rows and 1 column , row1 column 1 , y equals negative , 1 fourth , x squared , plus x plus 1 , row2 column 1 , y equals x minus 4 , end table
-
Business A company's weekly revenue R is given by the formula
R
=
−
p
2
+
30
p
,
r equals negative , p squared , plus 30 p comma where p is the price of the company's product. The company is considering hiring a distributor, which will cost the company
4
p
+
25
4 p plus 25 per week.
- Use a system of equations to find the values of the price p for which the product will still remain profitable if they hire this distributor.
- Which value of p will maximize the profit after including the distributor cost?
Solve each system.
-
{
y
=
5
x
2
+
9
x
+
4
y
=
−
5
x
+
3
left brace . table with 2 rows and 1 column , row1 column 1 , y equals 5 , x squared , plus 9 x plus 4 , row2 column 1 , y equals negative 5 x plus 3 , end table
-
{
y
=
−
7
x
2
−
9
x
+
6
y
=
1
2
x
+
11
left brace . table with 2 rows and 1 column , row1 column 1 , y equals negative 7 , x squared , minus 9 x plus 6 , row2 column 1 , y equals , 1 half , x plus 11 , end table
-
{
y
=
x
2
+
3
x
+
6
y
=
−
x
+
2
left brace . table with 2 rows and 1 column , row1 column 1 , y equals , x squared , plus 3 x plus 6 , row2 column 1 , y equals negative x plus 2 , end table
-
{
y
=
−
4
x
2
+
7
x
+
1
y
=
3
x
+
2
left brace . table with 2 rows and 1 column , row1 column 1 , y equals negative 4 , x squared , plus 7 x plus 1 , row2 column 1 , y equals 3 x plus 2 , end table
-
Reasoning Sketch the graphs of
y
=
2
x
2
+
4
x
−
5
y equals 2 , x squared , plus 4 x minus 5 and
y
=
x
2
+
2
x
−
3
.
y equals , x squared , plus 2 x minus 3 . Change these equations into inequalities so the system has solutions that comprise:
- two non-overlapping regions
- one bounded region
Solve the systems by graphing. For each system indicate one point in the solution set.
-
{
y
<
x
2
−
1
y
>
3
x
2
−
3
left brace . table with 2 rows and 1 column , row1 column 1 , y less than , x squared , minus 1 , row2 column 1 , y greater than 3 , x squared , minus 3 , end table
-
{
y
>
x
2
y
<
−
x
2
+
1
left brace . table with 2 rows and 1 column , row1 column 1 , y greater than , x squared , row2 column 1 , y less than negative , x squared , plus 1 , end table
-
{
y
>
(
x
−
3
)
2
+
4
y
<
−
(
x
−
3
)
2
+
5
left brace . table with 2 rows and 1 column , row1 column 1 , y greater than . open , x minus 3 , close squared . plus 4 , row2 column 1 , y less than negative . open , x minus 3 , close squared . plus 5 , end table
C Challenge
- Find a value of a for which the line y = x + a separates the parabolas
y
=
x
2
−
3
x
+
2
y equals , x squared , minus 3 x plus 2 and
y
=
−
x
2
+
8
x
−
15
.
y equals negative , x squared , plus 8 x minus 15 .
