Respuesta :
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)
