# express y=-3(x+1)^2-5 in standard form?

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- PopeLv 74 weeks ago
The expression "standard form" is a bit loaded. This is what I would call standard form for that parabola equation:

y = -3(x + 1)² - 5

-3(x + 1)² - 5 = y

-3(x + 1)² = y + 5

(x + 1)² = -1/3(y + 5) ... standard form

The curve has these parameters:

Vertex: (-1, -5)

Axis: x = -1

Latus rectum: 1/3

Downward concavity

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