Question

What is the the vertex of `y = -x^2 - 3x+12`?

Answer 1

`y = -(x+3/2)^2+57/4 " is in the form " y = a(x+b)^2 +c`

The vertex is at `(-3/2, +57/4)`

Explanation:

This is graph of an inverted parabola. To find the vertex (the maximum turning point) write the equation in the form

`y = acolor(blue)((x+b)^2) color(magenta)(+c)` which gives us the vertex as `(-b, c)`

We use the process of "completing the square"

Note:`(x-5)^2 = x^2 -10x+25 " in particular" (10/2)^2 = 25`

In the square of a binomial , this relationship is always true.

`color(red)((b/2)^2 )= c`

STEP 1 make the coefficient of `x^2` term = +1.

`y = -(1x^2 +3x -12) color(white)(xx)(-12 " is not the correct value for c")` .

STEP 2 add and subtract `color(red)((b/2)^2 )" which is " (3/2)^2`

`y = -[color(blue)(x^2 +3x +(3/2)^2) color(magenta)(-(3/2)^2-12])`

STEP 3 Write the first 3 terms as a perfect square

`y = -[color(blue)((x+3/2)^2) + color(magenta)((-(9/4)-48/4))]`

STEP 4 Simplify the constant

`y = -[ color(blue)((x+3/2)^2)color(magenta)(-57/4)]`

STEP 5 Remove the outer bracket to get vertex form.

`y = - color(blue)((x+3/2)^2)color(magenta)(+57/4)`

`y = acolor(blue)((x+b)^2) color(magenta)(+c)`

The vertex is at `(-3/2, +57/4)`

graph{-x^2 -3x+12 [-3.834, 1.166, 12.52, 15.02]}