Answer: Find the vertex form of h(x) = x2 - 14x + 6?
Solution:
It is given that,
h(x) = x2 - 14x + 6
We know that the vertex form is obtained by completing the square.
So we have to add and subtract a term, which is formed by the squared power of half the coefficient of the linear term:
= x2 - 14x + 6
= x2 - 14x + 6 + (14/2)2 - (14/2)2
= x2 - 14x + 6 + 72 - 72
= x2 - 14x + 72 + 6 - 72
Use the algebraic identity,
(a - b)2 = a2 - 2ab + b2
= (x - 7)2 + 6 - 72
= (x - 7)2 + 6 - 49
= (x - 7)2 - 43
Therefore, the vertex form of h(x) = x2 - 14x + 6 is (x - 7)2 - 43.
Find the vertex form of h(x) = x2 - 14x + 6?
Summary:
The vertex form of h(x) = x2 - 14x + 6 is (x - 7)2 - 43.
Explanation: