Given: A parabola equation
[tex]y=(x-3)^2-5[/tex]Required: To find the vertex of the given parabola.
Explanation: The given equation can be written as
[tex]\begin{gathered} y=x^2+9-6x-5 \\ y=x^2-6x+4 \end{gathered}[/tex]Now the general equation is of the form
[tex]y=ax^2+bx+c[/tex]Comparing both equations we get,
[tex]\begin{gathered} a=1 \\ b=-6 \\ c=4 \end{gathered}[/tex]The x-coordinate of the vertex is
[tex]\begin{gathered} x=-\frac{b}{2a} \\ x=\frac{-(-6)}{2(1)} \\ x=3 \end{gathered}[/tex]For the y coordinate of the vertex, put x=-3 in the equation of parabola
[tex]\begin{gathered} y=(3)^2-6(3)+4 \\ y=-5 \end{gathered}[/tex]Hence the coordinate of the vertex is (-3,31)
Final Answer: The vertex is (3,-5).
