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

1 Answer
Jul 12, 2017

#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.