Answer with explanation:
The given function : [tex]f(x)=x^2+2x-3[/tex]
Using completing the squares, we have
[tex]f(x)=x^2+2x+1-1-3[/tex] [∵ [tex](x+1)^2=x^2+2x+1[/tex]]
[tex]f(x)=(x+1)^2-4[/tex] (1)
Comparing (1) to the standard vertex form [tex]f(x)=(x-h)^2+k[/tex] , the vertex of function is at (h,k)=(-1,-4)
For x-intercept, put f(x)=0 in (1), we get
[tex]0=(x+1)^2-4\\\\\Rightarrow\ (x+1)^2=4[/tex]
Square root on both sides, we get
[tex]x+1=\pm2\\\\x+1=-2\ or\ \ x+1=2\\\\=x=-3\ \ or\ x=1[/tex]
∴ x intercepts : x= (-3,0) and (1,0)
For y-intercept put x=0 in (1), we get
[tex]f(1)=(1)^2-4=1-4=-3[/tex]
∴ y intercept : (0,-3)
Axis of symmetry : [tex]\dfrac{-b}{2a}[/tex]
In [tex]f(x)=x^2+2x-3[/tex] , a=1 and b=2
Axis of symmetry=[tex]\dfrac{-2}{2(1)}=-1[/tex]
