Question

The function f(x) = x2 − 12x + 5 written in vertex form is f(x) = (x − 6)2 − 31. What are the coordinates of the vertex?

Answer 1

$\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \boxed{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\ -------------------------------\\\\ f(x)=1(x-\stackrel{h}{6})^2\stackrel{k}{-31}\qquad \qquad vertex~(6,-31)$

Answer 2

Answer:-  C is the right answer. The coordinates of the vertex = (h,k) = (6, −31).

Explanation:-

Given standard form :- $f(x)=x^2-12x+5$

and its vertex form :-$f(x)=(x-6)^2-31$ [which derived from the completing the square form.]

On comparing with the vertex form of equation in parabola = $f(x)=(x-h)^2+k$

Then the coordinates of the vertex = (h,k) = (6, −31)

• A point (h,k) where a parabola intersects its axis of symmetry is called the vertex (h,k) of the parabola .