Resources

Resources

Bytelearn - cat image with glasses

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Identify the axis of symmetry, the vertex, and the
y-intercept of the graph. Then describe the end be
axis of symmetry:
◻
vertex:
(0,1)

y-intercept: 1
As
x increases or decreases,
y
◻ increases

Identify the axis of symmetry, the vertex, and the $y$ -intercept of the graph. Then describe the end be $\newline$ axis of symmetry: $\square$ $\newline$ vertex: $(0,1)$ $\newline$ $y$ -intercept: $1$ $\newline$ As $x$ increases or decreases, $y$ $\square$ increases

View step-by-step help

View step-by-step help

Reflections of functions

Full solution

Q. Identify the axis of symmetry, the vertex, and the

y

-intercept of the graph. Then describe the end be

\newline

axis of symmetry:

\square

\newline

vertex:

(0,1)

\newline

y

-intercept:

1

\newline

As

x

increases or decreases,

y

\square

increases

Parabola Axis of Symmetry: The axis of symmetry for a parabola given by the equation $y = ax^2 + bx + c$ is $x = -\frac{b}{2a}$ .
Given Vertex: For the given vertex $(0,1)$ , the axis of symmetry is $x = 0$ .
Finding Y-Intercept: The vertex is already given as $(0,1)$ .
Parabola Behavior: To find the $y$ -intercept, set $x$ to $0$ and solve for $y$ . Since the vertex is at $(0,1)$ , the $y$ -intercept is also $1$ .
Parabola Behavior: To find the $y$ -intercept, set $x$ to $0$ and solve for $y$ . Since the vertex is at $(0,1)$ , the $y$ -intercept is also $1$ .As $x$ increases or decreases from the axis of symmetry, the value of $y$ will increase since the parabola opens upwards ( $a > 0$ ).

More problems from Reflections of functions

Question

What does the transformation

f(x) \mapsto f(x) - 6

do to the graph of

f(x)

\newline

\newline

\text{translates it } 6 \text{ units down}

\newline

\text{translates it } 6 \text{ units right}

\newline

\text{translates it } 6 \text{ units left}

\newline

\text{translates it } 6 \text{ units up}

Posted 2 months ago

Question

g(x)

g(x)

is the reflection across the

y

f(x) = -4x + 1

\newline

\newline

\text{[g(x) = 4x - 1]}

\newline

\text{[g(x) = -4x + 1]}

\newline

\text{[g(x) = 4x + 1]}

\newline

\text{[g(x) = -4x - 1]}

Posted 2 months ago

Question

What kind of transformation converts the graph of

f(x) = -8(x - 4)^2 + 6

into the graph of

g(x) = -8(x + 6)^2 + 6

\newline

\newline

\text{[A]translation 10 units left}

\newline

\text{[B]translation 10 units right}

\newline

\text{[C]translation 10 units up}

\newline

\text{[D]translation 10 units down}

Posted 2 months ago

Question

g(x)

g(x)

is the reflection across the x-axis of

f(x) = -6(x - 7)^2 + 2

\newline

\newline

g(x) = 6(x - 7)^2 - 2

\newline

g(x) = -6(x + 7)^2 + 2

\newline

g(x) = -6(x - 7)^2 - 2

\newline

g(x) = 6(x + 7)^2 - 2

Posted 2 months ago

Question

\lim _{x \rightarrow \frac{\pi}{4}} \cos (x)=?

\newline

1

\newline

\frac{1}{2}

\newline

1

\newline

\frac{\sqrt{2}}{2}

\newline

(D) The limit doesn't exist.

Posted 3 months ago

Question

Tess tried to solve the differential equation

\frac{d y}{d x}=2 x y-6 x

. This is her work:

\newline

\frac{d y}{d x}=2 x y-6 x

\newline

1

\quad \frac{d y}{d x}=(y-3) 2 x

\newline

2

\quad \int \frac{1}{y-3} d y=\int 2 x d x

\newline

3

\quad \ln |y-3|=x^{2}+C_{1}

\newline

4

\quad e^{\ln |y-3|}=e^{x^{2}}+C_{1}

\newline

5

\quad|y-3|=e^{x^{2}}+C_{1}

\newline

6

\quad y-3= \pm e^{x^{2}}+C_{2}

\newline

7

\quad y= \pm e^{x^{2}}+C

\newline

Is Tess's work correct? If not, what is her mistake?

\newline

1

\newline

(A) Tess's work is correct.

\newline

1

is incorrect. We're not allowed to factor an expression when solving differential equations.

\newline

4

is incorrect. The right-hand side of the equation should be

e^{x^{2}+C_{1}}

\newline

7

is incorrect. Tess didn't account for adding

3

to the right side.

Posted 1 month ago

Question

Aditya tried to solve the differential equation

\frac{d y}{d x}=6 x-30

. This is his work:

\newline

\frac{d y}{d x}=6 x-30

\newline

1

\quad \int 30 d y=\int 6 x d x

\newline

2

\quad 30 y=3 x^{2}+C

\newline

3

\quad y=\frac{x^{2}}{10}+C

\newline

Is Aditya's work correct? If not, what is his mistake?

\newline

1

\newline

(A) Aditya's work is correct.

\newline

1

is incorrect. The separation of variables wasn't done correctly.

\newline

2

is incorrect. Aditya didn't integrate

30

\newline

3

is incorrect. The right-hand side of the equation should be

\frac{30}{3 x^{2}+C}

Posted 2 months ago

Question

Jonael dropped a sandbag from a hot air balloon at a target that is

250

meters below the balloon. At this moment, the sandbag is

75

meters below the balloon.

\newline

2

of the following expressions represents the distance between the sandbag and the target?

\newline

2

\newline

|-250+75|

\newline

|250+75|

\newline

|-75+250|

Posted 2 months ago

Question

f(x) = -(x+1)(x+7)

\newline

The function represents a parabola in the

xy

-plane. Which of the following is an equivalent form of

f

y

-intercept of the graph of

f

appears as a constant or coefficient?

\newline

1

\newline

f(x) = -(x+4)^{2}+9

\newline

f(x) = -(x+4)^{2}-9

\newline

f(x) = -x^{2}+8x+7

\newline

f(x) = -x^{2}-8x-7

Posted 2 months ago

Question

Determine whether the function

f(x)

is continuous at

x=-3

\newline

f(x)=\left\{\begin{array}{ll} 18-x^{2}, & x \leq-3 \\ 15+3 x, & x>-3 \end{array}\right.

\newline

f(x)

is discontinuous at

x=-3

\newline

f(x)

is continuous at

x=-3

Posted 2 months ago