Question

(02.05 MC)

Using the completing-the-square method, rewrite f(x) = x2 - 6x + 2 in vertex form.
Of(x) = (x - 3)2
Of(x) = (x - 3)2 + 2
Of(x) = (x - 3)2 - 7
Of(x) = (x - 3)2 + 9

Answer 1

`\\ \rm\hookrightarrow x^2-6x+2`

`\\ \rm\hookrightarrow x^2-2(3)(x)+2`

`\\ \rm\hookrightarrow x^2-2(3x)+3^2-3^2+2`

• 3^2-3^2=0

`\\ \rm\hookrightarrow (x-3)^2-9+2`

`\\ \rm\hookrightarrow (x-3)^2-7`

Answer 2

`f(x)=(x-3)^2-7`

Explanation:

Use the formula:

`x^2+bx+c \implies (x+(b)/(2))^2-((b)/(2){)^2+c`

`f(x) = x^2 - 6x + 2`

`\implies f(x)=(x-\frac62)^2-(\frac62)^2+2`

`\implies f(x)=(x-3)^2-3^2+2`

`\implies f(x)=(x-3)^2-9+2`

`\implies f(x)=(x-3)^2-7`