Question

What’s the vertex form of F(x)=x^2+2x-3

Answer 1

Answer:

$\large\boxed{f(x)=(x+1)^2-4}$

Step-by-step explanation:

$\text{The vertex form of a quadratic equation}\ f(x)=ax^2+bx+c:\\\\f(x)=a(x-h)^2+k\\\\(h,\ k)-\text{vertex}\\=====================================$

$\bold{METHOD\ 1:}\\\\\text{convert to the perfect square}\ (a+b)^2=a^2+2ab+b^2\qquad(*)\\\\f(x)=x^2+2x-3=\underbrace{x^2+2(x)(1)+1^2}_{(*)}-1^2-3\\\\f(x)=(x+1)^2-4\\==============================$

$\bold{METHOD\ 2:}\\\\\text{Use the formulas:}\ h=\dfrac{-b}{2a},\ k=f(k)\\\\f(x)=x^2+2x-3\to a=1,\ b=2,\ c=-3\\\\h=\dfrac{-2}{2(1)}=\dfrac{-2}{2}=-1\\\\k=f(-1)=(-1)^2+2(-1)-3=1-2-3=-4\\\\f(x)=(x-(-1))^2-4=(x+1)^2-4$