Question

determine the vertex of y= -x^2-7x-12

Answer 1

Two ways:

1) if you have a parabola in form \(y = ax^2+bx+c\), \(x\) value of the vertex is at \(x = -\dfrac{b}{2a}\) and plug that value of \(x\) into the formula to find the \(y\) value.

2) complete the square or otherwise rearrange the formula to put the parabola's equation into "vertex form", which is \(y = a(x-h)^2+k\) at which point you can read off the vertex's coordinates as \((h,k)\)

#1 should be easy enough, so I'll demonstrate #2

\[y = -x^2-7x-12 = -1(x^2+7x+12)\]We want to rewrite \((x^2+7x)\) as a perfect square. To do that, we take the coefficient of \(x\), which is 7, divide it by 2, and square it, giving us 49/4. That tells us that \((x+\frac{7}{2})^2 = x^2+7x+\frac{49}4\) is a perfect square, and all we have to do is figure out what is left after we take that away from \(x^2+7x+12\). \(12 = \frac{48}{4}\), so to do that we have

\[x^2+7x+\frac{48}{4} = (x^2+7x+\frac{49}4) +(\frac{48}{4}-\frac{49}{4})\]\[(x+7x+\frac{48}{4}) = (x+\frac{7}{2})^2-\frac{1}{4}\]We had to add and subtract 49/4 from the same side of the equation so we didn't actually change the value, just rearranged the pieces a bit for our convenience. Putting that back inside our original equation:
\[y = -x^2-7x-12 = -1(x^2+7x+12) = -1((x+\frac{7}2)^2-\frac{1}{4}) = -(x+\frac{7}2)^2+\frac{1}{4}\]gives us our desired "vertex form", and we can see that the parabola opens downward (because \(a < 0\), and our vertex is at \((-\dfrac{7}{2},\dfrac{1}{4})\). Remember, vertex form assumes we have \((x-h)^2\) and our value of \(h = -\frac{7}{2}\) because we had \((x+\frac{7}2)^2 = (x-h)^2\) or \(-h = \frac{7}{2}\)

Answer 2

genius!