Question

What would y=x^2 +x+ 2 be in vertex form

Answer 1

`y = (x + (1)/(2) )^(2) + (7)/(4)`

Explanation:

`y = {x}^(2) + x + 2`

We can covert the standard form into the vertex form by either using the formula, completing the square or with calculus.

`y = a(x - h)^(2) + k`

The following equation above is the vertex form of Quadratic Function.

Vertex — Formula

`h = - (b)/(2a) \\ k = \frac{4ac - {b}^(2) }{4a}`

We substitute the value of these terms from the standard form.

`y = a {x}^(2) + bx + c`

`h = - (1)/(2(1)) \\ h = - ( 1)/(2)`

Our h is - 1/2

`k = \frac{4(1)(2) - ( {1})^(2) }{4(1)} \\ k = (8 - 1)/(4) \\ k = (7)/(4)`

Our k is 7/4.

Vertex — Calculus

We can use differential or derivative to find the vertex as well.

`f(x) = a {x}^(n)`

Therefore our derivative of f(x) —

`f'(x) = n * a {x}^(n - 1)`

From the standard form of the given equation.

`y = {x}^(2) + x + 2`

Differentiate the following equation. We can use the dy/dx symbol instead of f'(x) or y'

`f'(x) = (2 * 1 {x}^(2 - 1) ) + (1 * {x}^(1 - 1) ) + 0`

Any constants that are differentiated will automatically become 0.

`f'(x) = 2 {x}+ 1`

Then we substitute f'(x) = 0

`0 =2x + 1 \\ 2x + 1 = 0 \\ 2x = - 1 \\x = - (1)/(2)`

Because x = h. Therefore, h = - 1/2

Then substitute x = -1/2 in the function (not differentiated function)

`y = {x}^(2) + x + 2`

`y = ( - (1)/(2) )^(2) + ( - (1)/(2) ) + 2 \\ y = (1)/(4) - (1)/(2) + 2 \\ y = (1)/(4) - (2)/(4) + (8)/(4) \\ y = (7)/(4)`

Because y = k. Our k is 7/4.

From the vertex form, our vertex is at (h,k)

Therefore, substitute h = -1/2 and k = 7/4 in the equation.

`y = a {(x - h)}^(2) + k \\ y = (x - ( - (1)/(2) ))^(2) + (7)/(4) \\ y = (x + (1)/(2) )^(2) + (7)/(4)`