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Genrish500 [490]
3 years ago
15

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

Mathematics
1 answer:
balu736 [363]3 years ago
7 0

Answer:

y = (x +  \frac{1}{2} )^{2}  +  \frac{7}{4}

Step-by-step 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.

<u>Vertex</u><u> </u><u>—</u><u> </u><u>Formula</u>

h =  -  \frac{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 =  -  \frac{1}{2(1)}  \\ h =  -  \frac{ 1}{2}

Our h is - 1/2

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

Our k is 7/4.

<u>Vertex</u><u> </u><u>—</u><u> </u><u>Calculus</u>

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 \times 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 \times 1 {x}^{2 - 1} ) + (1 \times  {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 =  -  \frac{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 = ( -  \frac{1}{2} )^{2}  + ( -  \frac{1}{2} ) + 2 \\ y =  \frac{1}{4}  -  \frac{1}{2}  + 2 \\ y =  \frac{1}{4}  -  \frac{2}{4}  +  \frac{8}{4}  \\ y =  \frac{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 - ( -  \frac{1}{2} ))^{2}  +  \frac{7}{4}  \\ y = (x +  \frac{1}{2} )^{2}  +  \frac{7}{4}

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