Answer:
We have the function, [tex]f(x)=(x+3)^{2}-4[/tex]
On simplifying, we get,
[tex]f(x)=x^2+6x+9-4[/tex]
i.e. [tex]f(x)=x^2+6x+5[/tex]
Thus, the standard form of the function is [tex]f(x)=x^2+6x+5[/tex].
Now, the factors of the given functions are (x+1) and (x+5).
Since, the intercept form of the function is the factored form of the function.
So, we have, intercept form of the function is [tex]f(x)=(x+1)(x+5)[/tex].
Now, we know that,
Value of x-coordinate of the vertex is [tex]\frac{-b}{2a}[/tex] i.e. [tex]\frac{-6}{2\times 1}[/tex] i.e. [tex]\frac{-6}{2}[/tex] i.e. x= -3
Then, [tex]f(-3)=(-3)^2+6\times (-3)+5[/tex] i.e. [tex]f(-3)=9-18+5[/tex] i.e. f(-3)=-4
So, the vertex of the function is (-3,-4).
Further, we know that, 'the y-intercept of a function is the point where the function crosses y-axis'.
So, when x=0, we have, [tex]f(0)=0^2+6\times 0+5[/tex] i.e. f(0) = 5
Thus, the y-intercept is (0,5)
Also, 'the x-intercept of a function is the point where the function crossese x-axis'.
Then, for f(x)=0, we have [tex]0=x^2+6x+5[/tex] i.e. [tex]0=(x+1)(x+5)[/tex] i.e. x= -1 and x= -5
Thus, the x-intercept are (-1,0) and (-5,0).
