Question

Why is f(x)=(3x+13)2+89 not the vertex form of f(x)=9x2+2x+1?

Answer 1

Explanation:

Given function in the vertex form is,

f(x) = (3x + 13)² + 89

=
`9(x+(13)/(3))^(2)+89` --------(1)

Vertex of the parabola →
`(-(13)/(3),89)`

If the standard equation of this function is,

f(x) = 9x² + 2x + 1

We will convert it into the vertex form,

f(x) = 9x² + 2x + 1

=
`9(x^(2)+(2)/(9)x)+1`

=
`9[x^(2)+2((1)/(9))x+((1)/(9))^(2)-((1)/(9))^2]+1`

=
`9[x^(2)+2((1)/(9))x+((1)/(9))^(2)]-9((1)/(9))^2+1`

=
`9[x^(2)+2((1)/(9))x+((1)/(9))^(2)]-((1)/(9))+1`

=
`9(x+(1)/(9))^2+(9-1)/(9)`

=
`9(x+(1)/(9))^2+(8)/(9)` -------(2)

Vertex of the function →
`(-(1)/(9),(8)/(9))`

Equation (1) and (2) are different and both the equations have different vertex.

Therefore, given equation doesn't match the equation given in the vertex form of the function.