Respuesta :
Solution
Gievn the equation below
[tex]4y-9=x^2-6x[/tex]To find the vertex and focus of the given equation, we apply the parabola standard equation which is
[tex]4p(y-k)=(x-h)^2[/tex]Where p is the focal length and the vertex is (h,k)
Rewriting the equation in standard form gives
[tex]\begin{gathered} 4y-9=x^2-6x \\ 4y=x^2-6x+9 \\ 4y=x^2-3x-3x+9 \\ 4y=x(x-3)-3(x-3) \\ 4y=(x-3)^2 \\ 4(1)(y-0)=(x-3)^2 \end{gathered}[/tex]Relating the parabola standard equation with the given equation, the vertex of the parabola is
[tex]\begin{gathered} x-3=0 \\ x=3 \\ y-0=0 \\ y=0 \\ (h,k)\Rightarrow(3,0) \\ p=1 \end{gathered}[/tex]Hence, the vertex is (3,0)
The focus of the parabola formula is
[tex](h,k+p)[/tex]Where
[tex]\begin{gathered} h=3 \\ k=0 \\ p=1 \end{gathered}[/tex]Substitute the values of h, k and p into the focus formula
[tex](h,k+p)\Rightarrow(3,0+1)\Rightarrow(3,1)[/tex]Hence, the focus is (3, 1)
