Originally Posted by

**gilagila** Write the following into the form of

$\displaystyle f(x)=a(x-h)^2+k$

and find the vertex, the symmetric axis and y-intercept

a) $\displaystyle f(x)=x^2+2x-1$

$\displaystyle y=x^2+2x-1$

$\displaystyle 1=x^2+2x$

$\displaystyle 1+1^2=x^2+2x+1^2$

$\displaystyle 2=(x+1)^2$

$\displaystyle y=(x+1)^2-2$

$\displaystyle f(x)=(x+1)^2-2$

Vertex = (-1,-2), Symmetric, x=-1, y-intercept, when x=0, y=-1. Q:Note this correction.

b) $\displaystyle f(x)=-x^2+4x-1$

Q:Your working is incorrect. $\displaystyle -x^2+4x+1\neq x^2-4x-1$

Starting again:

$\displaystyle -x^2+4x-1$

$\displaystyle =-[x^2-4x+1]$

$\displaystyle =-[(x-2)^2-2^2+1]$

$\displaystyle =-(x-2)^2+3$

Now solve for the vertices, symmetry etc.

same goes to

c) $\displaystyle f(x)=3x^2+6x-3$

Q:Again, incorrect. $\displaystyle \frac{f(x)}{3}\neq{f(x)}}$ - the only time you can divide through by a common factor is when identifying roots.

$\displaystyle f(x)=3[x^2+2x-1]$

$\displaystyle =3[(x+1)^2-1^2-1]$

$\displaystyle =3[(x+1)^2-2]$

$\displaystyle =3(x+1)^2-6$