Given:
[tex]f(x)=-(x+1)^2+4[/tex]
For "y" intercept of function tha value of x is zero.
[tex]\begin{gathered} f(x)=-(x+1)^2+4 \\ f(0)=-(0+1)^2+4 \\ =-1+4 \\ =3 \end{gathered}[/tex]
Y - intercept is 3.
For x inercept value of y is zero.
[tex]\begin{gathered} f(x)=-(x+1)^2+4 \\ -(x+1)^2+4=0 \\ (x+1)^2=4 \\ x+1=\pm2 \\ x=1,-3 \end{gathered}[/tex]
Vertex of function is
[tex]\begin{gathered} f^{\prime}(x)=0 \\ f(x)=-(x+1)^2+4 \\ f^{\prime}(x)=-2(x+1)_{} \\ -2(x+1)=0 \\ x=-1 \\ f(-1)=4 \\ \text{vertex}=(-1,4) \end{gathered}[/tex]
Graph of function is:
