Question

What is the range of the function $ y = 8x - 3 $ ?

Answer 1

Hint: We are given a function $ y = 8x - 3 $ . We have to find the range of this function. Range is the set of all the possible values that the function $ y = 8x - 3 $ can take. Here, the given equation is the equation of a straight line and we can easily find the range of a straight line. The range of this straight line is discussed below.

Complete step by step solution:
In this question, we are given a function $ y = 8x - 3 $ and we have to find the range of this function.
Now, range is the set of all the possible values that the function can take.
So, we have to find all the possible values that the function $ y = 8x - 3 $ can take.
Here, note that the given function $ y = 8x - 3 $ is the equation of straight line with slope 8 and y – intercept $ - 3 $ .
As we discussed earlier, the range of a function is the set of all valid outputs over its domain.
The domain of all the straight lines is $ \left( { - \infty , + \infty } \right) $ since they are defined for all values of x.
Hence, the domain of the given function
 $ y = 8x - 3 $ is $ \left( { - \infty , + \infty } \right) $ .
Now, since our function has no upper and lower bounds, the range of the given function
 $ y = 8x - 3 $ will be also $ \left( { - \infty , + \infty } \right) $ .
Therefore, the range of the function $ y = 8x - 3 $ is $ \left( { - \infty , + \infty } \right) $ .
So, the correct answer is “ $ \left( { - \infty , + \infty } \right) $ ”.

Note: We can also find the range of the given function using the method below.
Simplifying the given equation, we get
 $
   \Rightarrow y = 8x - 3 \\
   \Rightarrow 8x = y + 3 \\
   \Rightarrow x = \dfrac{{y + 3}}{8} \;
  $
Here, if $ y = - 3 $ , then $ x = 0 $ .
If $ y = - 11 $ , then $ x = - 1 $
If $ y = 5 $ , then $ x = 1 $
Here, Observe that y can take all the real values and hence the range of $ y = 8x - 3 $ will be equal to $ \left( { - \infty , + \infty } \right) $ .