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

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lisboa lisboa · Answer 1 · 2018-10-25T07:22:45+02:00

the vertex of the quadratic function

[tex]f(x)= \frac{1}{2}x^2+3x+\frac{3}{2}[/tex]

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

[tex]x = \frac{-b}{2a}[/tex]

[tex]x= \frac{-3}{2\frac{1}{2}}=-3[/tex]

x coordinate of vertex is -3

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

[tex]f(-3)= \frac{1}{2}(-3)^2+3(-3)+\frac{3}{2}[/tex]

f(-3) = -3

the y coordinate of vertex is -3

So vertex is (-3,-3)

alinakincsem alinakincsem · Answer 2 · 2018-10-28T19:18:34+01:00

Answer:

(-3, -3)

Step-by-step explanation:

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

Here, [tex]a= \frac{1}{2} , b = 3[/tex] and [tex]c= \frac{3}{2}[/tex].

We know the formula to find the x coordinate = [tex]\frac{-b}{2a}[/tex]

[tex]x= \frac{-3}{2(\frac{1}{2}) }[/tex] = [tex]-3[/tex]

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

[tex]y = \frac{1}{2} x^2+3x+\frac{3}{2}[/tex]

[tex]y = \frac{1}{2} (-3)^2+3(-3)+\frac{3}{2}[/tex]

[tex]y = -3[/tex]

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