Question 921157
The vertex form of a parabola is given by  y = a(x - h)^2 + k, a ≠ 0. 
standard form is y=  ax^2+bx+c 
Comparing with the standard form of parabola, f(x) = ax^2 + bx + c, we get
a = 4. 
           
     given equation   y= 4x^2+24x-13 
                      y = (2x)^2 -2*2x*6-13
                 to make it perfect square  add 6^2 and subtract 6^2
                y = (2x)^2 -2*2x*6-13+6^2-6^2

               y=(2x)^2 -2*2x*6+6^2 -13-6^2
               y= (2x-6)^2-13-36
                y =(2x-6)^2 -49
                  take 2 out side in square term
               y = 2^2 *( x-3)^2 -49

               y= 4 (x-3)^2-49