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

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Answer 1 · 2017-07-12T22:05:00

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

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.

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