Question

How to find the vertex of 3(x-1)^2 -5

Answer 1

`\sf\\Let\ y=3(x-1)^2-5\\or,\ y=3(x^2-2x+1)-5\\or,\ y=3x^2-6x+3-5\\or,\ y=3x^2-6x-2.......(1)\\\textsf{Comparing with y = }ax^2+bx+c,\ a=3,\ b=-6\ and\ c=-2\\\textsf{The equation of line of symmetry of vertical parabola is given by:}\\\\x=-(b)/(2a)=-(-6)/(2(3))=1\\\\\textsf{So x = 1 is the line of symmetry of the parabola.}`

`\sf\\\textsf{Now, vertex is the point where the line of symmetry and parabola intersects.}\\\textsf{Putting x = 1 in equation (1),}\\y=3(1)^2-6(1)-2\\or,\ y=3-6-2\\or,\ y=-5\\\textsf{So the vertex of parabola is (1,-5).}`