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# Find the vertex form of h(x) = x^{2} - 14x + 6?

**Solution:**

It is given that,

h(x) = x^{2} - 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:**

= x^{2} - 14x + 6

= x^{2} - 14x + 6 + (14/2)^{2} - (14/2)^{2}

= x^{2} - 14x + 6 + 7^{2} - 7^{2}

= x^{2} - 14x + 7^{2} + 6 - 7^{2}

**Use the algebraic identity,**

**(a - b) ^{2} = a^{2} - 2ab + b^{2}**

= (x - 7)^{2} + 6 - 7^{2}

= (x - 7)^{2} + 6 - 49

= (x - 7)^{2} - 43

**Therefore, the vertex form of h(x) = x ^{2} - 14x + 6 is (x - 7)^{2} - 43.**

## Find the vertex form of h(x) = x^{2} - 14x + 6?

**Summary:**

The vertex form of h(x) = x^{2} - 14x + 6 is (x - 7)^{2} - 43.

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