Question

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

Answer 1

f(x) = (x - $\frac{11}{2}$)² - $\frac{85}{4}$

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

$x_{vertex}$ = - $\frac{b}{2a}$

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

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

$x_{vertex}$ = - $\frac{-11}{2}$ = $\frac{11}{2}$

substitute this value into the equation for y- coordinate

y = ($\frac{11}{2}$)² - 11($\frac{11}{2}$) + 9

  = $\frac{121}{4}$ - $\frac{242}{4}$ + $\frac{36}{4}$ = - $\frac{85}{4}$

f(x) = (x - $\frac{11}{2}$)² - $\frac{85}{4}$ ← in vertex form