Please can you help with ths math involving vertex form?

For this quadratic equation


a)express the relation in factored form


b)find the coordinates of the vertex


c)express the relation in vertex form





y=4x^2 - 12 + 5





Thanks in advance.

Please can you help with ths math involving vertex form?
Hi,





For y=4x^2 - 12x + 5, it factors into y = (2x - 5)(2x - 1)


That means its x intercepts are the answers to 2x - 5 = 0 and 2x - 1 = 0., wjhich would be x = 5/2 and x = 1/2.





For the vertex of y=4x^2 - 12x + 5, find the axis of symmetry by doing x = -b/(2a). That would be x = -(-12)/(2*4) = 12/8 or 3/2 Note that this is the average of the 2 x intercepts too, showing they are equally spaced to each side. If you plug 3/2 in for x, you can solve for its y value of -4. Your vertex is the point (3/2, -4).





To write y=4x^2 - 12x + 5 in vertex form, first factor a 4 ( the x^2 coefficient) out of both x terms.





y=4(x^2 - 3x.........) + 5 Now complete the square in the parentheses. 1/2 of -3 is -3/2. -3/2 squared is 9/4 so add 9/4 inside the parentheses to complete the square.





y=4(x^2 - 3x.+ 9/4) + 5 To find out how much that changed the equation, you must multiply the 4 out front times the 9/4 to get 9. That means the equation overall is 9 more than it used to be, so to balance that out, you have to subtract 9 out on the end behind the 5. Then combine like terms.





y=4(x^2 - 3x + 9/4) + 5 - 9


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





Now factor the parentheses into a binomial squared. It will always start with x, have the same sign as what is in front of the x term, and the number you had earlier before you squared it.





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





This is vertex form, y = a(x - h)^2 + k, where (h, k) is the vertex. Note it's the same as what we found in part 2.








I hope that helps!! :-)
Reply:a) (2x-5)(2x-1) = y


b) Take the derivative


dy/dx = 8x -12 = 0, set = 0 to find vertex


x = 12/8 = 3/2


c) I do not know what vertex form is, sorry