Question

Put the quadratic into vertex form and state the coordinates of the vertex y = x² - 6x + 45

Answer 1

`\begin{gathered} y=(x-3)^2-(-3)^2+45=(x-3)^2-9+45=(x-3)^2+36 \\ \Rightarrow y-36=(x-3)^2 \end{gathered}`

Alternatively

Vertex coordinate =(h, k)

h= -b/2a

k = f(h)

From x^2 - 6x + 45

b= -6, a= 1

`h\text{ = }(-b)/(2a)\text{ =}\frac{-(-6)_{}}{2(1)}\text{ =}(6)/(2)=\text{ 3}`

k = f(h)

Substitute h in x^2 -6x +45

`\begin{gathered} k=f(3)=3^2\text{ - 6(3) + 45} \\ =\text{ 9-18 +45} \\ =\text{ 54-18=36} \end{gathered}`

The vertex coordinate is ( 3, 36)

The vertex form is

`y=(x-3)^2\text{ + 36}`