Question

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

Answer 1

`y = 1/4(x-2)^2-5`

Explanation:

The given equation

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

is in standard form:

`y = ax^2+ bx + c`

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

Here is a graph of the given equation:

graph{x^2/4 - x - 4 [-8.55, 11.45, -6.72, 3.28]}

The vertex form for a parabola of this type is:

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

where `(h,k)` is the vertex.

We know that "a" in the standard form is the same as the vertex form, therefore, we substitute `1/4` for "a" into equation [2]:

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

To find the value of `h`, we use the formula:

`h = -b/(2a)`

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

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

`h = 2`

Substitute 2 for `h` into equation [3]:

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

To find the value of k, we evaluate the given equation at `x = h = 2`:

`k= (2)^2/4 - 2 - 4`

`k = 1 - 2 - 4`

`k = -5`

Substitute -5 for `k` into equation [4]:

`y = 1/4(x-2)^2-5`

Here is a graph of the vertex form:

graph{1/4(x-2)^2-5 [-8.55, 11.45, -6.72, 3.28]}

Please observe that the two graphs are identical.