# How to write the quadratic x^2-14x+50 to the vertex form?

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### 2 Answers

`x^2-14x+50 `

`(x^2-14x)+50 `

use `(-b/2)^2`

`-14/2=-7^2=49 `

add it in the parentheses and minus it on the outside

`(x^2-14x+49)+50-49`

`(x^2-14x+49)+1 `

find the square root of a and c

`(x-7)^2+1 `

the x is 7 and the y is the +1 on the outside

We'll recall the vertex form of the equation of a parabola:

f(x) = a(x-h)^2 + k

"a" represents the leading coefficient: a = 1

(h,k) are the coordinate of the vertex of parabola.

We'll create a perfect square within the given quadratic:

f(x) = (x^2 - 14x + 49) + 1

f(x) = (x - 7)^2 + 1

The coordinates of the vertex of the parabola are (7 , 1).

**The vertex form of the given quadratic f(x) = x^2 - 14x + 50 is f(x) = (x - 7)^2 + 1.**