Question

How do you write `f(x) = 2x^2+x− 6` in vertex form?

Answer 1

By factoring, using the "complete the square" method.

Explanation:

Given the polynomial equation

`f(x)=2x^2+x-6`

Step 1: Factor out the `x^2` coefficient (in this case, `2`) from both the `x^2` and `x` terms:

`color(white)(f(x))=2(x^2+1/2 x)-6`

Step 2: Complete the square by taking the new `x` coefficient (in this case, `1/2`), dividing it in half (`1/4`), then squaring this value (`1/16`), and adding and subtracting this squared constant inside the bracket:

`color(white)(f(x))=2(x^2+1/2 x+1/16-1/16)-6`

(The `+1/16` will make the trinomial a perfect square; the `-1/16` keeps the value of the expression the same.)

Step 3: Move the subtracted constant (`-1/16`) outside of the bracket by multiplying it by the brackets' coefficient, if there is one (in this case, `2`):

`color(white)(f(x))=2(x^2+1/2 x+1/16)-6-1/8`

Step 4: The trinomial inside the brackets is a perfect square; it is the square of `(x+"''half of the "x" coefficient''")` (in this case, `(x+1/4)`). Rewrite the trinomial as this square, and combine the constant terms outside (`-6` and `-1/8`):

`color(white)(f(x))=2(x+1/4)^2-49/8`

And there it is! The equation is now in vertex form, and the vertex of the parabola is `(–1/4," "–49/8)`. The `x`-coordinate is the `1/4` but with the opposite sign; the `y`-coordinate is just the `-49/8`.