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The function
y=g(x) where
g(x)=3((1)/(2))^(-x)+2 is graphed in the
xy-plane. Which of the following is a true statement?
Choose 1 answer:
(A) The graph of function
g is always increasing.
(B) The
y-intercept of the graph of function
g is
(0,2).
(C) The
x-intercept of the graph of function
g is
(0,3).
(D) Function
g is symmetric with respect to the
y-axis.

The function $y=g(x)$ where $g(x)=3\left(\frac{1}{2}\right)^{-x}+2$ is graphed in the $x y$ -plane. Which of the following is a true statement? $\newline$ Choose $1$ answer: $\newline$ (A) The graph of function $g$ is always increasing. $\newline$ (B) The $y$ -intercept of the graph of function $g$ is $(0,2)$ . $\newline$ (C) The $x$ -intercept of the graph of function $g$ is $(0,3)$ . $\newline$ (D) Function $g$ is symmetric with respect to the $y$ -axis.

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Dilations of functions

Full solution

Q. The function

y=g(x)

where

g(x)=3\left(\frac{1}{2}\right)^{-x}+2

is graphed in the

x y

-plane. Which of the following is a true statement?

\newline

Choose

1

answer:

\newline

(A) The graph of function

g

is always increasing.

\newline

(B) The

y

-intercept of the graph of function

g

is

(0,2)

.

\newline

(C) The

x

-intercept of the graph of function

g

is

(0,3)

.

\newline

(D) Function

g

is symmetric with respect to the

y

-axis.

Check Function Growth: Examine the function $g(x) = 3(\frac{1}{2})^{-x} + 2$ to see if it's always increasing. $\newline$ The base $(\frac{1}{2})$ is less than $1$ , and the exponent is negative, which makes the function an exponential growth function.
Find Y-Intercept: Check the y-intercept by plugging in $x = 0$ into $g(x)$ . $g(0) = 3(\frac{1}{2})^{(-0)} + 2 = 3(1) + 2 = 3 + 2 = 5$ . The y-intercept is $(0, 5)$ , not $(0, 2)$ .
Locate X-Intercept: Look for the x-intercept by setting $g(x)$ to $0$ and solving for $x$ . $\newline$ $0 = 3(\frac{1}{2})^{-x} + 2$ . $\newline$ Subtract $2$ from both sides: $-2 = 3(\frac{1}{2})^{-x}$ . $\newline$ This equation has no solution because $3(\frac{1}{2})^{-x}$ is always positive, so there's no x-intercept.
Test Symmetry: Check if the function $g$ is symmetric with respect to the $y$ -axis by testing if $g(-x) = g(x)$ . $g(-x) = 3(\frac{1}{2})^{-(-x)} + 2 = 3(\frac{1}{2})^x + 2$ , which is not equal to $g(x)$ .Therefore, the function is not symmetric with respect to the $y$ -axis.

More problems from Dilations 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