y=4x^2+3x-1

Question

I want to ask what is the range equation. And it's domain and range.

Answer 1

The function y = 4x ^2 + 3x - 1.

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

Center (h, k ).

Here x ² coefficient is 4, for perfect square make x ² coefficient 1 by deviding each side by 4.

y / 4 = x ^2 + (3 / 4)x - (1 / 4).

To change the expression into a perfect square trinomial add (half the x coefficient)² to each side of the expression.

x  coefficient = 3 / 4 then (half the x coefficient)² is (3 / 8)^2 = 9 / 64.

So, Add 9 / 64 to each side.

y / 4 + 9 / 64 = x ^2 + (3 / 4)x + 9 / 64 - (1 / 4)

y / 4 + 9 / 64 = [x + (3 / 8)]^2 - (1 / 4)

y / 4  = [x + (3 / 8)]^2 - (1 / 4) - 9 / 64

y / 4  = [x + (3 / 8)]^2 - 25 / 64

y = 4[x + (3 / 8)]^2 - 25 / 16.

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

 a  = 4 , h  = - 3 / 8 and k  = - 25 / 16.

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 (- 3/8, - 25/16).

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  = - 25 / 16  so the graph of parabola cannot be lower than - 25 / 16 .

Thus the range of function y  ≥ - 25 / 16.

Domain of function is all real numbers.

Range of the function is  {y |y  ≥ - 25 / 16}.