How do I find the domain and range of this function:

Question

f(x)=(-2x-4)^2 - 3 ?? Please help! :)?

Answer 1

y = f(x).

y = (-2x-4)^2 - 3

Suppose, let x = 0

y =(-2*0-4)^2 - 3

y = 16-3

y = 13

Similarly, x =1

y = (-2*1-4)^2 -3

y = 36-3

y = 33

Let x =-1

y =(-2*-1 - 4)^2 - 3

y = 4-3

y = 1

Let x =2

y =(-2*2 -4)^2 - 3

y = 64-3

y = 61

Let x =-2

y =(-2*-2 -4)^2 -3

y =-3

If we let x be -2,-1,0,1,2 respectively and substitute each value of x in to_

the function,then the set of ordrs pairs would be { (-2,-3), (-1,1), (0,13), (1,33), (2,61) }

Domain is x values and range is y values.

So the domain set ={..,-2,-1,0,1,2,...}

Range = {...,-3,1,13,33,61,...}                                                   (for all real numbers)

Answer 2

The function f (x ) = (- 2x - 4)² - 3

y  = (- 2x - 4)² - 3

We first put the equation in to the form for a translated parabola y = a (x - h )² + k .

Center (h, k ).

In the next step we factored 4 from the right hand side to make the coefficient of our x  "+1" as this is the standard form.

y = 4(x + 2)² - 3

y = 4(x -(-2))² - 3

The above function represents a parabola vertex form  y = a (x - h )² + k .

 a  = 4 , h  = -2 and k  = -3.

a  is positive number the parabola opens up and has minimum value.

When the parabola opens up it has a minimum point which is the vertex of parabola (-2, -3)

We know that domain of the function is all possible x  values and range is all posible y  values.

 parabola domain x  =  all real numbers.

In the minimum point y  = -3  so the graph of parabola cannot be lower than -3.

Thus the range of function y  ≥ -3.

Domain of function is all real numbers.

Range of the function is  .