The points on the graph represent both an exponential function and a linear function.

complete this table by reading the values from the graph. estimate any function values that are less than one.
x -3 -2 -1 0 1 2 3
exponential function
linear function
at approximately what values of x do both the linear and exponential functions have the same value for y?

By looking at the graphs we can infer the equation that represents both functions. If we plot the equations in a graphing tool, we can easily find your points.

Please, see the attached image

Linear

f(x)  = (-3/2)*x + 3

x = -3 ---------> y = 15/2 = 7.5

x = -2 ---------> y = 6

x = -1 ---------> y = 4.5

x = 0 ---------> y = 3

x = 1 ---------> y = 1.5

x = 2 ---------> y = 0

x = 3 ---------> y = -1.5

Exponential

g(x)  = (1/2)^x

x = -3 ---------> y = 8

x = -2 ---------> y = 4

x = -1 ---------> y = 2

x = 0 ---------> y = 1

x = 1 ---------> y = 0.5

x = 2 ---------> y = 0.25

x = 3 ---------> y = 0.125

(1/2)^x =  (-3/2)*x + 3

If we solve this equation, we get two values

x = -2.868 and

x = 1.81

Step-by-step explanation:

From the graph attached we will find the linear as well as exponential functions first.

Afterwards we will plug in the values of x to get the value of the function given.

For Linear function

It should be in the form of y = mx + c

We find c = 3

and two points passing through the line are (0, 3) and (2, 0)

so slope of the line should be

Now we can say the linear function becomes

For x = -3,

For x = -2

For x = -1

For x = 0

For x = 1

For x = 2 f(2) = -6

For x = 3

Now for Exponential function

function will be in the form of

Since point (-1, 2) is passing through the exponential function

So

⇒

Therefore exponential function is

Now from the given graph

g(-3) = 8

g(-2) = 6

g(-1) = 2

g(0) = 1

Now we will try to get the common values of x by analyzing the graphs of two functions.

we get the solutions for x as (-2.87) and (1.81)

x+14

step-by-step explanation:

i think the answer is b.

step-by-step explanation:

not 100% sure though.