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Let’s check out your problem:
Find the
y
y
y
-coordinate of the
y
y
y
-intercept of the polynomial function defined below.
\newline
f
(
x
)
=
4
x
−
4
x
2
+
5
x
5
+
3
x
3
+
5
x
4
f(x)=4 x-4 x^{2}+5 x^{5}+3 x^{3}+5 x^{4}
f
(
x
)
=
4
x
−
4
x
2
+
5
x
5
+
3
x
3
+
5
x
4
\newline
Answer:
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Home
Math Problems
Algebra 1
Reflections of functions
Full solution
Q.
Find the
y
y
y
-coordinate of the
y
y
y
-intercept of the polynomial function defined below.
\newline
f
(
x
)
=
4
x
−
4
x
2
+
5
x
5
+
3
x
3
+
5
x
4
f(x)=4 x-4 x^{2}+5 x^{5}+3 x^{3}+5 x^{4}
f
(
x
)
=
4
x
−
4
x
2
+
5
x
5
+
3
x
3
+
5
x
4
\newline
Answer:
Evaluate
f
(
x
)
f(x)
f
(
x
)
at
x
=
0
x=0
x
=
0
:
To find the
y
y
y
-coordinate of the
y
y
y
-intercept of the function
f
(
x
)
f(x)
f
(
x
)
, we need to evaluate
f
(
x
)
f(x)
f
(
x
)
when
x
=
0
x = 0
x
=
0
.
Substitute
x
=
0
x=0
x
=
0
into
f
(
x
)
f(x)
f
(
x
)
:
Substitute
x
=
0
x = 0
x
=
0
into the polynomial function
f
(
x
)
=
4
x
−
4
x
2
+
5
x
5
+
3
x
3
+
5
x
4
f(x) = 4x - 4x^2 + 5x^5 + 3x^3 + 5x^4
f
(
x
)
=
4
x
−
4
x
2
+
5
x
5
+
3
x
3
+
5
x
4
.
Calculate
f
(
0
)
f(0)
f
(
0
)
:
f
(
0
)
=
4
(
0
)
−
4
(
0
)
2
+
5
(
0
)
5
+
3
(
0
)
3
+
5
(
0
)
4
f(0) = 4(0) - 4(0)^2 + 5(0)^5 + 3(0)^3 + 5(0)^4
f
(
0
)
=
4
(
0
)
−
4
(
0
)
2
+
5
(
0
)
5
+
3
(
0
)
3
+
5
(
0
)
4
.
Simplify the expression:
Simplify the expression by calculating the value of each term when
x
=
0
x = 0
x
=
0
.
f
(
0
)
=
0
−
0
+
0
+
0
+
0.
f(0) = 0 - 0 + 0 + 0 + 0.
f
(
0
)
=
0
−
0
+
0
+
0
+
0.
Final result:
The result of
f
(
0
)
f(0)
f
(
0
)
is
0
0
0
, which means the y-coordinate of the
y-intercept
is
0
0
0
.
More problems from Reflections of functions
Question
What does the transformation
f
(
x
)
↦
f
(
x
)
−
6
f(x) \mapsto f(x) - 6
f
(
x
)
↦
f
(
x
)
−
6
do to the graph of
f
(
x
)
f(x)
f
(
x
)
?
\newline
Choices:
\newline
(A)
translates it
6
units down
\text{translates it } 6 \text{ units down}
translates it
6
units down
\newline
(B)
translates it
6
units right
\text{translates it } 6 \text{ units right}
translates it
6
units right
\newline
(C)
translates it
6
units left
\text{translates it } 6 \text{ units left}
translates it
6
units left
\newline
(D)
translates it
6
units up
\text{translates it } 6 \text{ units up}
translates it
6
units up
Get tutor help
Posted 9 months ago
Question
Find
g
(
x
)
g(x)
g
(
x
)
, where
g
(
x
)
g(x)
g
(
x
)
is the reflection across the
y
y
y
-axis of
f
(
x
)
=
−
4
x
+
1
f(x) = -4x + 1
f
(
x
)
=
−
4
x
+
1
.
\newline
Choices:
\newline
[g(x) = 4x - 1]
\text{[g(x) = 4x - 1]}
[g(x) = 4x - 1]
\newline
[g(x) = -4x + 1]
\text{[g(x) = -4x + 1]}
[g(x) = -4x + 1]
\newline
[g(x) = 4x + 1]
\text{[g(x) = 4x + 1]}
[g(x) = 4x + 1]
\newline
[g(x) = -4x - 1]
\text{[g(x) = -4x - 1]}
[g(x) = -4x - 1]
Get tutor help
Posted 9 months ago
Question
What kind of transformation converts the graph of
f
(
x
)
=
−
8
(
x
−
4
)
2
+
6
f(x) = -8(x - 4)^2 + 6
f
(
x
)
=
−
8
(
x
−
4
)
2
+
6
into the graph of
g
(
x
)
=
−
8
(
x
+
6
)
2
+
6
g(x) = -8(x + 6)^2 + 6
g
(
x
)
=
−
8
(
x
+
6
)
2
+
6
?
\newline
Choices:
\newline
[A]translation 10 units left
\text{[A]translation 10 units left}
[A]translation 10 units left
\newline
[B]translation 10 units right
\text{[B]translation 10 units right}
[B]translation 10 units right
\newline
[C]translation 10 units up
\text{[C]translation 10 units up}
[C]translation 10 units up
\newline
[D]translation 10 units down
\text{[D]translation 10 units down}
[D]translation 10 units down
Get tutor help
Posted 9 months ago
Question
Find
g
(
x
)
g(x)
g
(
x
)
, where
g
(
x
)
g(x)
g
(
x
)
is the reflection across the x-axis of
f
(
x
)
=
−
6
(
x
−
7
)
2
+
2
f(x) = -6(x - 7)^2 + 2
f
(
x
)
=
−
6
(
x
−
7
)
2
+
2
.
\newline
Choices:
\newline
(A)
g
(
x
)
=
6
(
x
−
7
)
2
−
2
g(x) = 6(x - 7)^2 - 2
g
(
x
)
=
6
(
x
−
7
)
2
−
2
\newline
(B)
g
(
x
)
=
−
6
(
x
+
7
)
2
+
2
g(x) = -6(x + 7)^2 + 2
g
(
x
)
=
−
6
(
x
+
7
)
2
+
2
\newline
(C)
g
(
x
)
=
−
6
(
x
−
7
)
2
−
2
g(x) = -6(x - 7)^2 - 2
g
(
x
)
=
−
6
(
x
−
7
)
2
−
2
\newline
(D)
g
(
x
)
=
6
(
x
+
7
)
2
−
2
g(x) = 6(x + 7)^2 - 2
g
(
x
)
=
6
(
x
+
7
)
2
−
2
Get tutor help
Posted 9 months ago
Question
lim
x
→
π
4
cos
(
x
)
=
?
\lim _{x \rightarrow \frac{\pi}{4}} \cos (x)=?
x
→
4
π
lim
cos
(
x
)
=
?
\newline
Choose
1
1
1
answer:
\newline
(A)
1
2
\frac{1}{2}
2
1
\newline
(B)
1
1
1
\newline
(C)
2
2
\frac{\sqrt{2}}{2}
2
2
\newline
(D) The limit doesn't exist.
Get tutor help
Posted 1 year ago
Question
Tess tried to solve the differential equation
d
y
d
x
=
2
x
y
−
6
x
\frac{d y}{d x}=2 x y-6 x
d
x
d
y
=
2
x
y
−
6
x
. This is her work:
\newline
d
y
d
x
=
2
x
y
−
6
x
\frac{d y}{d x}=2 x y-6 x
d
x
d
y
=
2
x
y
−
6
x
\newline
Step
1
1
1
:
d
y
d
x
=
(
y
−
3
)
2
x
\quad \frac{d y}{d x}=(y-3) 2 x
d
x
d
y
=
(
y
−
3
)
2
x
\newline
Step
2
2
2
:
∫
1
y
−
3
d
y
=
∫
2
x
d
x
\quad \int \frac{1}{y-3} d y=\int 2 x d x
∫
y
−
3
1
d
y
=
∫
2
x
d
x
\newline
Step
3
3
3
:
ln
∣
y
−
3
∣
=
x
2
+
C
1
\quad \ln |y-3|=x^{2}+C_{1}
ln
∣
y
−
3∣
=
x
2
+
C
1
\newline
Step
4
4
4
:
e
ln
∣
y
−
3
∣
=
e
x
2
+
C
1
\quad e^{\ln |y-3|}=e^{x^{2}}+C_{1}
e
l
n
∣
y
−
3∣
=
e
x
2
+
C
1
\newline
Step
5
5
5
:
∣
y
−
3
∣
=
e
x
2
+
C
1
\quad|y-3|=e^{x^{2}}+C_{1}
∣
y
−
3∣
=
e
x
2
+
C
1
\newline
Step
6
6
6
:
y
−
3
=
±
e
x
2
+
C
2
\quad y-3= \pm e^{x^{2}}+C_{2}
y
−
3
=
±
e
x
2
+
C
2
\newline
Step
7
7
7
:
y
=
±
e
x
2
+
C
\quad y= \pm e^{x^{2}}+C
y
=
±
e
x
2
+
C
\newline
Is Tess's work correct? If not, what is her mistake?
\newline
Choose
1
1
1
answer:
\newline
(A) Tess's work is correct.
\newline
(B) Step
1
1
1
is incorrect. We're not allowed to factor an expression when solving differential equations.
\newline
(C) Step
4
4
4
is incorrect. The right-hand side of the equation should be
e
x
2
+
C
1
e^{x^{2}+C_{1}}
e
x
2
+
C
1
.
\newline
(D) Step
7
7
7
is incorrect. Tess didn't account for adding
3
3
3
to the right side.
Get tutor help
Posted 1 year ago
Question
Aditya tried to solve the differential equation
d
y
d
x
=
6
x
−
30
\frac{d y}{d x}=6 x-30
d
x
d
y
=
6
x
−
30
. This is his work:
\newline
d
y
d
x
=
6
x
−
30
\frac{d y}{d x}=6 x-30
d
x
d
y
=
6
x
−
30
\newline
Step
1
1
1
:
∫
30
d
y
=
∫
6
x
d
x
\quad \int 30 d y=\int 6 x d x
∫
30
d
y
=
∫
6
x
d
x
\newline
Step
2
2
2
:
30
y
=
3
x
2
+
C
\quad 30 y=3 x^{2}+C
30
y
=
3
x
2
+
C
\newline
Step
3
3
3
:
y
=
x
2
10
+
C
\quad y=\frac{x^{2}}{10}+C
y
=
10
x
2
+
C
\newline
Is Aditya's work correct? If not, what is his mistake?
\newline
Choose
1
1
1
answer:
\newline
(A) Aditya's work is correct.
\newline
(B) Step
1
1
1
is incorrect. The separation of variables wasn't done correctly.
\newline
(C) Step
2
2
2
is incorrect. Aditya didn't integrate
30
30
30
correctly.
\newline
(D) Step
3
3
3
is incorrect. The right-hand side of the equation should be
30
3
x
2
+
C
\frac{30}{3 x^{2}+C}
3
x
2
+
C
30
.
Get tutor help
Posted 1 year ago
Question
Jonael dropped a sandbag from a hot air balloon at a target that is
250
250
250
meters below the balloon. At this moment, the sandbag is
75
75
75
meters below the balloon.
\newline
Which
2
2
2
of the following expressions represents the distance between the sandbag and the target?
\newline
Choose
2
2
2
answers:
\newline
(A)
∣
−
250
+
75
∣
|-250+75|
∣
−
250
+
75∣
\newline
(B)
∣
250
+
75
∣
|250+75|
∣250
+
75∣
\newline
(C)
∣
−
75
+
250
∣
|-75+250|
∣
−
75
+
250∣
Get tutor help
Posted 1 year ago
Question
f
(
x
)
=
−
(
x
+
1
)
(
x
+
7
)
f(x) = -(x+1)(x+7)
f
(
x
)
=
−
(
x
+
1
)
(
x
+
7
)
\newline
The function represents a parabola in the
x
y
xy
x
y
-plane. Which of the following is an equivalent form of
f
f
f
in which the
y
y
y
-intercept of the graph of
f
f
f
appears as a constant or coefficient?
\newline
Choose
1
1
1
answer:
\newline
(A)
f
(
x
)
=
−
(
x
+
4
)
2
+
9
f(x) = -(x+4)^{2}+9
f
(
x
)
=
−
(
x
+
4
)
2
+
9
\newline
(B)
f
(
x
)
=
−
(
x
+
4
)
2
−
9
f(x) = -(x+4)^{2}-9
f
(
x
)
=
−
(
x
+
4
)
2
−
9
\newline
(C)
f
(
x
)
=
−
x
2
+
8
x
+
7
f(x) = -x^{2}+8x+7
f
(
x
)
=
−
x
2
+
8
x
+
7
\newline
(D)
f
(
x
)
=
−
x
2
−
8
x
−
7
f(x) = -x^{2}-8x-7
f
(
x
)
=
−
x
2
−
8
x
−
7
Get tutor help
Posted 1 year ago
Question
Determine whether the function
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
−
3
x=-3
x
=
−
3
.
\newline
f
(
x
)
=
{
18
−
x
2
,
x
≤
−
3
15
+
3
x
,
x
>
−
3
f(x)=\left\{\begin{array}{ll} 18-x^{2}, & x \leq-3 \\ 15+3 x, & x>-3 \end{array}\right.
f
(
x
)
=
{
18
−
x
2
,
15
+
3
x
,
x
≤
−
3
x
>
−
3
\newline
f
(
x
)
f(x)
f
(
x
)
is discontinuous at
x
=
−
3
x=-3
x
=
−
3
\newline
f
(
x
)
f(x)
f
(
x
)
is continuous at
x
=
−
3
x=-3
x
=
−
3
Get tutor help
Posted 1 year ago
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