Determine the vertex and axis of symmetry of the following quadratics. a) f(x)=2(x−3)2−1 b) f(x)=x2+4x+3

Step 1
For a quadratic equation ${\displaystyle y=a{x}^2+bx+c}$
Axis of Symmetry is ${\displaystyle x=\frac{-b}{2a}}$
x-coordinate of vertex is ${\displaystyle \frac{-b}{2a}}$
a) ${\displaystyle f\left(x\right)=2{(x-3)}^2-1}$
${\displaystyle =2(x^2-6x+9)-1}$
${\displaystyle f\left(x\right)=2x^2-12x+17}$
Axis of symmetry ${\displaystyle \Rightarrow x=\frac{-b}{2a}}$
${\displaystyle \Rightarrow x=-\frac{(-12)}{2\left(2\right)}=3}$
${\displaystyle \Rightarrow x=3}$
Step 2
x-coordinate of vertex ${\displaystyle =\frac{-b}{2a}=3}$
y-coordinate of vertex ${\displaystyle =f\left(\frac{-b}{2a}\right)=f\left(3\right)}$
${\displaystyle =2{(3-3)}^2-1=-1}$
${\displaystyle \therefore }$ Vertex ${\displaystyle =(3,-1)}$
b) ${\displaystyle f\left(x\right)=x^2+4x+3}$
Axis of symmetry
${\displaystyle \Rightarrow x=\frac{-b}{2a}=\frac{-4}{2\left(1\right)}=-2}$
${\displaystyle \Rightarrow x=-2}$
x-coordinate of vertex ${\displaystyle =\frac{-b}{2a}=-2}$
Step 3
y-coordinate of vertex
${\displaystyle =f\left(\frac{-b}{2a}\right)}$
${\displaystyle =f(-2)}$
${\displaystyle ={(-2)}^2+4(-2)+3}$
${\displaystyle =-1}$
${\displaystyle \therefore }$ Vertex ${\displaystyle =(-2,-1)}$

Step 1
${\displaystyle f\left(x\right)=2(x-3)2-1}$
Multiply 2 and 2 to get 4.
${\displaystyle 4(x-3)-1}$
Use the distributive property to multiply 4 by ${\displaystyle x-3}$.
${\displaystyle 4x-12-1}$
Subtract 1 from -12 to get -13
${\displaystyle 4x-13}$
Step 2
${\displaystyle f\left(x\right)=x^2+4x+3}$
Factor the expression by grouping. First, the expression needs to be rewritten as ${\displaystyle x^2+ax+bx+3}$. To find a and b, set up a system to be solved
${\displaystyle a+b=4}$
${\displaystyle ab=1\times 3=3}$
Since ab is positive, a and b have the same sign. Since ${\displaystyle a+b}$ is positive, a and b are both positive. The only such pair is the system solution.
${\displaystyle a=1}$
${\displaystyle b=3}$
Rewrite ${\displaystyle x^2+4x+3}$ as ${\displaystyle (x^2+x)+(3x+3)}$
${\displaystyle (x^2+x)+(3x+3)}$
Factor out x in the first and 3 in the second group.
${\displaystyle x(x+1)+3(x+1)}$
Factor out common term ${\displaystyle x+1}$ by using distributive property.
${\displaystyle (x+1)(x+3)}$