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MrRa [10]
3 years ago
13

Which graph models the function f(x)=-2(3)^x

Mathematics
2 answers:
Vitek1552 [10]3 years ago
8 0

Answer:

Reflection across x-axis:

(x, y) \rightarrow (x, -y)

Vertically stretch:

A function y=a f(x) is vertically stretch by a factor a > 1 is that of parent function y=f(x)

Given the graph: f(x) = -2(3)^x

We will make a table values for a few values of x.

then we will graph the given function.

x                f(x)

-2            -0.2222..

-1             -0.6666..

0                 -2

1                  -6

2                 -18

3                 -54

Note that as x increases, f(x) decreases

Now, using these points (-2, -0.2222..), (-1, -0.666..), (0, -2), (1, -6) , (2, -18) and (3, -54)

Plot the graph of the given function as shown below:

We observe that the curve is that of f(x) =(3)^x except it is vertically stretch by a factor 2 and reflection across x-axis.

svetlana [45]3 years ago
4 0

Solution:

\text{Given: }f(x)=-2(3)^x

We are an exponential function.

y=ab^x

If a<0 and b>1 then graph is decreasing.

If a>0 and b<1 then graph is decreasing.

If a<0 and b<1 then graph is increasing.

If a>0 and b>1 then graph is increasing.

\text{We have a function }f(x)=-2(3)^x

Here a=-2<0 and b=3>1 therefore, f(x) is decreasing.

Horizontal asymptote , y=0

x-intercept: Doesn't not exist

y-intercept: (0,-2)

Using above information we will draw the graph f(x)

We make table of x and y for different value of x

  x        y

-2       -0.22

 -1       -0.67

 0          -2

 1           -6

 2           -18

Plot these points on graph and join the points.

 Please see the attachment to see the graph.



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<h3><u>Option A</u></h3>

4 x^{2}+32 x-80=0 is the required equation to calculate width of rectangular frame that has a total area of 140 square inches.

<h3><u>Solution:</u></h3>

Given that,

Length of a rectangular frame is given as 2x + 10

Width of the rectangular frame is given as 2x + 6

Total area = 140 square inches

<em><u>The area of rectangular frame is given as:</u></em>

\text {area of rectangle}=\text {length } \times \text {width}

Plugging in values, we get

\begin{array}{l}{(2 x+10)(2 x+6)=140} \\\\ {4 x^{2}+12 x+20 x=80} \\\\ {4 x^{2}+32 x-80=0}\end{array}

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