Question

I need the answers… I’ve asked like 3 different tutors and no one can help me so

Answer 1

Given:

`f(x)=-(x+1)^2+4`

For "y" intercept of function tha value of x is zero.

`\begin{gathered} f(x)=-(x+1)^2+4 \\ f(0)=-(0+1)^2+4 \\ =-1+4 \\ =3 \end{gathered}`

Y - intercept is 3.

For x inercept value of y is zero.

`\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}`

Vertex of function is

`\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}`

Graph of function is: