How do you find the vertex of 2x^2+2x+10

f(x) has a global minimum at x = -1/2

Step-by-step explanation:

Find and classify the global extrema of the following function:

f(x) = 2 x^2 + 2 x + 10

Find the critical points of f(x):

Compute the critical points of 2 x^2 + 2 x + 10

To find all critical points, first compute f'(x):

d/( dx)(2 x^2 + 2 x + 10) = 4 x + 2

= 2 (2 x + 1):

f'(x) = 2 (2 x + 1)

Solving 2 (2 x + 1) = 0 yields x = -1/2:

x = -1/2

f'(x) exists everywhere:

2 (2 x + 1) exists everywhere

The only critical point of 2 x^2 + 2 x + 10 is at x = -1/2:

x = -1/2

The domain of 2 x^2 + 2 x + 10 is R:

The endpoints of R are x = -∞ and ∞

Evaluate 2 x^2 + 2 x + 10 at x = -∞, -1/2 and ∞:

The open endpoints of the domain are marked in gray

x | f(x)

-∞ | ∞

-1/2 | 19/2

∞ | ∞

The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:

The open endpoints of the domain are marked in gray

x | f(x) | extrema type

-∞ | ∞ | global max

-1/2 | 19/2 | global min

∞ | ∞ | global max

Remove the points x = -∞ and ∞ from the table

These cannot be global extrema, as the value of f(x) here is never achieved:

x | f(x) | extrema type

-1/2 | 19/2 | global min

f(x) = 2 x^2 + 2 x + 10 has one global minimum:

 f(x) has a global minimum at x = -1/2

(-1/2, 9 1/2)

Step-by-step explanation:

Looking at 2x^2 + 2x + 10, we see that the coefficients are 2, 2 and 10.

The axis of symmetry is x = -b/(2a), which here comes out to

      -2

x = = -1/2.

      2(2)  

To determine the vertex, evaluate the given function at x = -1/2:

2(-1/2)^2 + 2(-1/2) + 10, or 2(1/4) - 1 + 10, or 1/2 + 9, or y = 9 1/2.

The vertex is at (-1/2, f(-1/2)), or (-1/2, 9 1/2).

answer: sine

step-by-step explanation: