How do you determine the vertex of this quadratic function?

Question

Graph and then choose the correct plot of the quadratic function.

f(x)=1/3 x^2 - 4/3

Standard form is f(x) = a(x - h)^2 + k

This parabola opens upward since the leading coefficient (a) is positive.

y-intercept = - 4/3
x-intercept = 2

How do you determine the vertex if it's not stated. I'm getting a little confused...but is the vertex (0, - 4/3)?

Thanks.

Answer 1

Vertex is at (h,k).

Since h = 0 and k = -4/3 you are correct about the vertex.

However, there are two x-intercepts: +2,-2

Answer 2

x= -b/2a* gives the x-coordinate of the vertex. and the y coordinate is just the function evaluated at that x value.

so, you where right, (0,-4/3) is the vertex.

*(b is the coefficient of the x term, here 0. a is the coefficient of the x^2 term, here 1/3)

Answer 3

f(x) = (1/3)x^2 - (4/3)

Obvious

h = 0
k = (-4/3)

-----------------------------------

f(x) = (1/3)x^2 - (4/3)

x = (-b)/(2a)

x = (-0)/(2(1/3))
x = 0/(2/3)
x = 0

f(0) = (1/3)(0)^2 - (4/3)
f(0) = (-4/3)

ANS :

(0,(-4/3))

x = (-b)/(2a) helps to find the max or min of the x value, then you just plug that in for x, to get the max or min of the y value.

y = a(-b/(2a))^2 + b(-b/(2a)) + c
y = ((ab^2)/(4a^2)) - ((b^2)/(2a)) + c
y = ((b^2)/(4a)) - ((b^2)/(2a)) + c
y = ((b^2)/(4a)) - ((2b^2)/(4a)) + ((4ac)/(4a))
y = (b^2 - 2b^2 + 4ac)/(4a)
y = (-b^2 - 4ac)/(4a)

so

x = (-b)/(2a)
y = -(b^2 + 4ac)/(4a)