How do you find the vertex of $f (x) = - 3 x^{2} - 6 x - 7$ $f(x)= -3x^2-6x-7$ ?

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Answer 1 · 2015-06-14T14:56:19

Complete the square to find:

$f (x) = - 3 x^{2} - 6 x - 7 = - 3 {(x + 1)}^{2} - 4$ $f(x) = -3x^2-6x-7 = -3(x+1)^2-4$

has vertex $(- 1, - 4)$ $(-1, -4)$

$f (x) = - 3 x^{2} - 6 x - 7$ $f(x) = -3x^2-6x-7$

$= - 3 x^{2} - 6 x - 3 - 4$ $= -3x^2-6x-3-4$

$= - 3 (x^{2} + 2 x + 1) - 4$ $= -3(x^2+2x+1)-4$

$= - 3 {(x + 1)}^{2} - 4$ $= -3(x+1)^2 - 4$

Now ${(x + 1)}^{2} \geq 0$ $(x+1)^2 >= 0$ has it's minimum value $0$ $0$ when $x = - 1$ $x = -1$

If $x = - 1$ $x=-1$ then

$f (x) = {(x + 1)}^{2} - 4 = 0^{2} - 4 = 0 - 4 = - 4$ $f(x) = (x+1)^2-4 = 0^2-4 = 0-4 = -4$

So the vertex is at $(- 1, - 4)$ $(-1, -4)$

In general you can complete the square of any quadratic in $x$ $x$ as follows:

$a x^{2} + b x + c = a {(x + \frac{b}{2 a})}^{2} + (c - \frac{b^{2}}{4 a})$ $ax^2+bx+c = a(x+b/(2a))^2 + (c-b^2/(4a))$

How do you find the vertex of f(x)= -3x^2-6x-7f(x)=−3x2−6x−7 f(x)= -3x^2-6x-7?