Find the vertex form of the following quadratic function. f(x)= x^2-11x+9

f(x) = (x - )² -

the equation of a quadratic in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

given a quadratic in standard form : y = ax² + bx + c ( a ≠ 0 )

then the x- coordinate of the vertex is

= -

f(x) = x² - 11x + 9 is in standard form

with a = 1, b = - 11 and c = 9

= - =

substitute this value into the equation for y- coordinate

y = ()² - 11() + 9

  = - + = -

f(x) = (x - )² - ← in vertex form

b.an increase in a causes b to decrease

is this part of a question or something