Given: The function below
[tex]y=-x^2+6[/tex]To Determine: The graph of the parabola and the vertex
Solution
Determine the vertex
The vertex form of a parabola is
[tex]\begin{gathered} y=a(x-h)^2+k \\ vertex=(h,k) \end{gathered}[/tex][tex]\begin{gathered} h=-\frac{b}{2a} \\ k=f(-\frac{b}{2a}) \end{gathered}[/tex]Note:
[tex]\begin{gathered} If \\ ax^2+bx+c=0 \\ y=-x^2+6 \\ a=-1,b=0,c=6 \end{gathered}[/tex][tex]\begin{gathered} Therefore \\ h=\frac{-(0)}{2\times-1}=0 \\ y=-x^2+6 \\ when,x=0 \\ y=-0^2+6 \\ y=0+6 \\ y=6 \end{gathered}[/tex]The coordinate of the vertex is (0, 6)
Let us take two points on the left
[tex]\begin{gathered} x=-2 \\ y=-(-2)^2+6 \\ y=-4+6 \\ y=2,(-2,2) \\ x=-1 \\ y=-(-1)^2+6 \\ y=-1+6 \\ y=5(-1,5) \end{gathered}[/tex]Let us take two points on the right
[tex]\begin{gathered} x=1 \\ y=-(1)^2+6 \\ y=-1+6 \\ y=5,(1,5) \\ x=2 \\ y=-(2)^2+6 \\ y=-4+6 \\ y=2,(2,2) \end{gathered}[/tex]Let us plot the 5 points as a graph as shown below
Hence, the vertex is (0, 6) and two points on the left of the vertex are (-1,5) and (-2, 2), and two points to the right of the vertex are (1, 5) and (2, 2)
