How do you find the axis of symmetry and the vertex for $y = -x^2 -10x?$

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Answer 1 · 2017-03-24T07:21:52

Axis of symmetry: $x = -5$

The vertex is at $(-10, 25)$

There is a formula to find the axis of symmetry of a parabola.
The standard form of a parabola is $y = ax^2 +bx +c$

Axis of symmetry: $x = (-b)/(2a)$

We have $y = -x^2 -10x$

$a = -1 and b= -10$

$x = (-(-10))/(2(-1)) = 10/-2 = -5$

$x=-5$ is the line of symmetry.

The vertex lies on the line of symmetry, so as soon as you have the $x$ -value, you can find the $y$ -value.

$y= -(-5)^2 -10(-5)$

$y= -25+50$

$y =25$

The vertex is at $(-10, 25)$

How do you find the axis of symmetry and the vertex for y = -x^2 -10x?y = -x^2 -10x?