Question

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

Answer 1

Answer: $(-3,-3)$

Step-by-step explanation:

Given the function:

$f(x)=\frac{1}{2}x^2+3x+\frac{3}{2}$

You can rewrite it:

$y=\frac{1}{2}x^2+3x+\frac{3}{2}$

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

$x=\frac{-b}{2a}$

In this case:

$a=\frac{1}{2}\\\\b=3$

Then:

 $x=\frac{-3}{2(\frac{1}{2})}=-3$

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

$y=\frac{1}{2}(-3)^2+3(-3)+\frac{3}{2}=-3$

Therefore, the vertex of the function is:

$(-3,-3)$

Answer 2

Answer:

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

Step-by-step 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)