What is the the vertex of #y = (x -3)^2+4x-5 #?

1 Answer
Jan 6, 2016

The solution set(or vertex set) is: #S = {-5,-21}.#

Explanation:

The standard formula of the quadratic function is:
#y = Ax^2 + Bx +C#

#(x-3)^2# is a notable product, so do this:
Square the first number -(signal inside the parenthesis) 2 * first number * second number + second number squared
#x^2 - 6x + 9#

Now, substitute it the main equation:
#y = x^2 - 6x + 9 +4x - 5 = x^2 +10x +4#, so
#y = x^2 +10x +4# #to# Now, it agrees with the standard formula.

To find the point of the vertex in #x# axis, we apply this formula:
#x_(vertex) = -b/(2a) = -10/2 = -5#

To find the point of the vertex in #y# axis, we apply this formula:
#y_(vertex) = - triangle/(4a) = - (b^2 - 4ac)/(4a) = -(100 -4 * 1 * 4)/4 = -21#

Then, the solution set(or vertex set) is: #S = {-5,-21}.#