Question

What is the vertex of the function f(x) = one-halfx2 + 3x + three-halves?

Answer 1

Answer:
`(-3,-3)`

Explanation:

Given the function:

`f(x)=(1)/(2)x^2+3x+(3)/(2)`

You can rewrite it:

`y=(1)/(2)x^2+3x+(3)/(2)`

You can find the x-coordinate of the vertex with this formula:

`x=(-b)/(2a)`

In this case:

`a=(1)/(2)\\\\b=3`

Then:

`x=(-3)/(2((1)/(2)))=-3`

Substituting "x" into the function, you get that the y-coordinate of the vertex is:

`y=(1)/(2)(-3)^2+3(-3)+(3)/(2)=-3`

Therefore, the vertex of the function is:

`(-3,-3)`

Answer 2

The vertex of the function is (-3 , -3)

Explanation:

* Lets explain how to find the vertex of the quadratic function

- The general form of the quadratic function is f(x) = ax² + bx + c,

where a , b , c are constant

- The vertex of the quadratic function is (h , k) , where

h = -b/2a and k = f(h)

* Lets solve the problem

∵ f(x) = one-half x² + 3x + three-halves

∴ f(x) = 1/2 x² + 3x + 3/2

∵ f(x) = ax² + bx + c

∴ a = 1/2 , b = 3 , c = 3/2

∵ The coordinates of its vertex is (h , k)

∵ h = -b/2a

∴ h = -3/2(1/2) = -3/1 = -3

∴ h = -3

∵ k = f(h)

∴ k = f(-3)

∵ f(-3) = 1/2 (-3)² + 3(-3) + 3/2 = -3

∴ k = -3

∴ The vertex of the function is (-3 , -3)