The discontinuity occurs at x = 0, since that is the only "problem" place in the graph that makes the function undefined. A vertical asymptote exists there. It is nonremoveable.
(x + 3)^2 + (x + 4)^2
= x^2 + 6x + 9 + x^2 + 8x + 16
= 2x^2 + 14x + 25
= 2(x^2 + 7x) + 25
= 2[(x + 7/2)^2 - 49/4] + 25
= 2(x + 7/2)^2 - 98/4 + 25
= 2(x + 7/2)^2 + 1/2
Its B
Problem:
Solve
2
x
+
5
y
=
−
3
;
3
x
−
y
=
21
Steps:
I will try to solve your system of equations.
3
x
−
y
=
21
;
2
x
+
5
y
=
−
3
Step: Solve
3
x
−
y
=
21
for y:
3
x
−
y
+
−
3
x
=
21
+
−
3
x
(Add -3x to both sides)
−
y
=
−
3
x
+
21
−
y
−
1
=
−
3
x
+
21
−
1
(Divide both sides by -1)
y
=
3
x
−
21
Step: Substitute
3
x
−
21
for
y
in
2
x
+
5
y
=
−
3
:
2
x
+
5
y
=
−
3
2
x
+
5
(
3
x
−
21
)
=
−
3
17
x
−
105
=
−
3
(Simplify both sides of the equation)
17
x
−
105
+
105
=
−
3
+
105
(Add 105 to both sides)
17
x
=
102
17
x
17
=
102
17
(Divide both sides by 17)
x
=
6
Step: Substitute
6
for
x
in
y
=
3
x
−
21
:
y
=
3
x
−
21
y
=
(
3
)
(
6
)
−
21
y
=
−
3
(Simplify both sides of the equation)
Answer:
y
=
−
3
and
x
=
6
