Answer:
Vertex: ( 4, -1)
Explanation:
The vertex form of a parabola is given by
[tex]f\mleft(x)=a\left(x-h\right)^2+k\mright)[/tex]
where is the location of the vertex is (h, k).
Now in our case. we have
[tex]f\mleft(x\mright)=-\left(x-4\right)^2-1[/tex]
From the above equation we recognize h = 4 and k = - 1; therefore, the vertex is
[tex]\left(4,-1\right)[/tex]
which is our answer!
Let us now find the x-intercepts.
To find the x-intercepts, we solve the following.
[tex]0=-\left(x-4\right)^2-1[/tex]
The first step is to add + 1 to both sides. This gives
[tex]1=-(x-4)^2[/tex]
The next step is to multiply both sides by - 1.
[tex]-1=(x-4)^2[/tex]
Then we take the square root of both sides. This gives
[tex]\sqrt{-1}=\sqrt{\left(x-4\right)^2}[/tex]
On the left, we see that we are taking the sqaure root of a negative number. This cannot be done since it gives imaginary ( and not real) numbers.
Hence, we conclude that the solutions to 0 = -(x - 4)^2 - 1 do not exist, and therefore, the parabola has no x-intercepts.
To find the y-intercept, we put x = 0 into our equation. This gives
[tex]y=-\left(0-4\right?^2-1[/tex][tex]y=-16-1[/tex][tex]\boxed{y=-17.}[/tex]
