How to determine the coordinates of the vertex of parabola, v(h,k), if y=x^2-6x+8.
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
The equation of the parabola is y = x^2 - 6x + 8
y = x^2 - 6x + 8
=> x^2 - 6x + 9 - 1
=> (x - 3)^2 -1
This is the standard form of a parabola y = (x - h)^2 + k
The vertex is at ( 3, -1)
giorgiana1976 | College Teacher | (Level 3) Valedictorian
We'll write the function as:
f(x) = a(x-h)^2 + k, where the vertex has the coordinates v(h,k)
We'll write the given function:
f(x) = 1(x^2 - 6x) + 8
We'll complete the square x^2 - 6x:
x^2 -2*(3)*x + (3)^2 = (x - 3)^2
So, we'll add and subtract the value 9:
f(x) = 1(x^2 - 6x + 9) - 9 + 8
f(x) = (x - 3)^2 - 1
We'll compare the result with the standard form:
(x - 3)^2 - 1= a(x-h)^2 + k
h = 3
k = -1
The coordinates of the vertex of parabola are:V (3 ; -1).