What is the vertex of the function x2 - 6x + 8 = 0?

Answer: The vertex of the function [tex]\bold{x^{2} -6 x + 8 = 0}[/tex]is (h,k) = (3 , -1) Solution: The vertex form of quadratic equation is generally given as, [tex]f(x) = a(x - h)^{2} + k[/tex]Where h,k is the vertex of the parabola. From question, given that [tex]x^{2}-6 x+8=0[/tex] . we have to find the vertex of the function. Let us first convert the given quadratic equation to vertex form (eqn 1) [tex]x^{2}-6 x+8=0[/tex] [tex]x^{2}-6 x=-8[/tex]By adding “9” on both sides of equation, we get [tex]x^{2}-6 x+9=-8+9[/tex][tex]x^{2}-6 x+9=1[/tex]By using the identity [tex](a-b)^{2}=a^{2}-2 a b+b^{2}[/tex] ,the right hand side of above equation becomes, [tex](x-3)^{2}=1[/tex][tex](x-3)^{2}-1=0[/tex]Now,the equation [tex](x-3)^{2}-1=0[/tex] is of the vertex form. By comparing [tex](x-3)^{2}-1=0[/tex] with [tex]a(x-h)^{2}+k[/tex]we get the values of (h,k) a = 1; h = 3; k = -1 hence the vertex of the function [tex]x^{2}-6 x+8=0[/tex] is (h,k) = (3 , -1)