Question

What is the vertex of the quadratic function f(x)=1/2x^2+3x+3/2?

Answer 1

(-3, -3)

Explanation:

The standard form of a quadratic equation is
`y = ax^2+bx+c` where the coordinates of the highest point (x, y) indicate the vertex.

Here,
`a= (1)/(2) , b = 3` and
`c= (3)/(2)`.

We know the formula to find the x coordinate =
`(-b)/(2a)`

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

To find y, put this value of x in the function to get:

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

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

`y = -3`

Therefore, the vertex (x, y) = (-3, -3)

Answer 2

the vertex of the quadratic function

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

To find vertex we use formula x= -b/2a

from the given equation , a= 1/2 and b = 3

Now we plug in the values in the formula

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

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

x coordinate of vertex is -3

Now plug in -3 for x in f(x)

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

f(-3) = -3

the y coordinate of vertex is -3

So vertex is (-3,-3)