Question #27abc

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Answer 1 · 2017-02-24T07:05:35

The vertex is at $(1, - 6)$ $(1,-6)$

There are different methods you can use.

A quadratic has the form $a x^{2} + b x + c$ $ax^2 +bx +c$

$y = x^{2} - 2 x - 5$ $y = x^2 -2x-5$ is the equation of a parabola.

The vertex lies on the line of symmetry of the parabola.
It is a vertical line and can be found from $x = \frac{- b}{2 a}$ $x = (-b)/(2a)$

Once you know the $x$ $x$ -value, substitute it into the equation of the parabola to find the $y$ $y$ -value.

$x = \frac{- b}{2 a} = \frac{- (- 2)}{2 (1)} = \frac{2}{2} = 1$ $x = (-b)/(2a) = (-(-2))/(2(1)) = 2/2 =1$

Now find $y$ $y$

$y = x^{2} - 2 x - 5 = {(1)}^{2} - 2 (1) - 5 = - 6$ $y = x^2 -2x-5 = (1)^2 -2(1)-5 = -6$

The vertex is at $(1, - 6)$ $(1,-6)$

OR:
Completing the square will give the vertex form.

OR:
You can draw the graph by finding points and plotting them and then read the vertex from the points or from the graph.

graph{y=x^2-2x-5 [-9.9, 10.1, -6.56, 3.44]}