Question

What is the vertex form of `y=1/3x^2+1/4x-1`?

Answer 1

`y = 1/3(x-(-3/8))^2-67/64 larr` this is the vertex form.

Explanation:

The given equation :

`y=1/3x^2+1/4x-1" [1]"`

Is in the standard form:

`y = ax^2+ bx + c" [2]"`

where `a =1/3, b = 1/4, and c = -1`

The desired vertex form is:

`y = a(x-h)^2+k" [3]"`

The "a" in equation [2] is the same value as the "a" in equation [3], therefore, we make that substitution:

`y = 1/3(x-h)^2+k" [4]"`

The x coordinate of the vertex, h, can found be using the values of "a" and "b" and the formula:

`h = -b/(2a)`

Substituting in the values for "a" and "b":

`h = -(1/4)/(2(1/3))`

`h = -3/8`

Substitute the value for h into equation [4]:

`y = 1/3(x-(-3/8))^2+k" [5]"`

The y coordinate of the vertex, k, can be found by evaluating equation [1] at `x = h = -3/8`

`k = 1/3(-3/8)^2+1/4(-3/8)-1`

`k = -67/64`

Substitute the value for k into equation [5]:

`y = 1/3(x-(-3/8))^2-67/64 larr` this is the vertex form.