3 years ago

What’s the vertex of #8 ? Please help!!!!

Mathematics

1 answer:

ANTONII [103]3 years ago

6 0

$y = x^2 + 2x + 1$

It's already a square; we don't need to add anything to complete the square.

$y = (x + 1)^2$

That's a vertex at x = -1. The corresponding y is y=0.

Answer: Vertex at (-1,0)

As for the other questions cut off there, the axis is the vertical line x = -1 and the range is y ≥ 0.

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Step-by-step explanation:

$Solve:\\\ 6(2x^2-5)x=-3\\\\Use\ distributive\ property\ first\\\\(12x^2-30)x=-3\\\\Now\ multiply\ the\ terms\ in\ ()\ by\ 'x'\\\\12x^3-30x=-3\\\\\large\boxed{12x^3-30x+3=0}$