Does (1,6) make the inequality x<1 true
1 answer:
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a) Increasing function.
b) The growth factor is 4.
c) The asymptote is y tending to zero as x tends to negative infinity.
d) y = 1
c) The graph can be seen at the end.
<h3>What conclusions can we get about the given function?</h3>Here we are given the following exponential function:
y =f(x) = 4^x
a) First, we want to know if the function is increasing or decreasing. Remember that a function is increasing if, as x increses, also does the value of y.
Here clearly we have an increasing function.
- f(0) = 4^0 = 1
- f(1) = 4^1 = 4
- f(2) = 4^2 = 16
and so on...
b) Now we want to identify the growth factor. This is the number that is being exponentiated, so in:
y = 4^x
The growth factor is 4.
c) We have an asymptote as x tends to negative infinity, because we will eventually have denominator that tends to infinity, then we will have an asymptote that tends to zero, as x tends to negative infinity.
d) The y-intercept is what we get when we evaluate in x = 0, we have:
y = f(0) = 4^0 = 1
The y-intercept is y = 1.
e) The sketch of the exponential function can be seen in the image below.
If you want to learn more about exponential functions:
#SPJ1
Answer:
a= -6
Step-by-step explanation:
Please see attached picture for full solution.
I skipped a few steps while rewriting the equation into the form of y=mx+c
Here's the step-by-step working if you don't understand:
2x -3y= 6
2x= 3y +6 (+3y on both sides)
2x -6= 3y (-6 on both sides)
3y= 2x -6
<em>y= 2/3x -2</em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>(÷3 throughout)
