How do I solve this algebra 2 problem: 4(y+2) = 3(x-1)^2

4(y+2) = 3(x-1)^2

So I need to write it in standard form then state whether it would make a parabola circle or ellipse.

if its a parabola i need to find:
vertex, axis of symetry, direction of opening

if its a circle:
center and radius

ellipse:
center major, minor axis lengths, foci coordinates

So I tried to solve it, but in the end I ended up confused. I will grant 10 points to the best, most explained, yet easy to understand answer.

Thanks! from Jen4197

Hi,

4(y + 2) = 3(x - 1)²

y + 2 = 3/4(x - 1)²

y = 3/4(x - 1)² - 2

Since "a" = 3/4, the graph opens up. <==ANSWER
The vertex ix (1,-2) and the axis of symmetry is x = 1. <==ANSWER

I hope that helps!! :-)

4(y+2) = 3(x-1)^2
4(y+2) = 3(x-1)(x-1)
4y+8 = 3(x^2-2x+1)
4y+8 = 3x^2-6x+3
4y = 3x^2-6x-5 removed 8 from both sides
y = 0.75x^2-1.5x-1.25 divided both sides by 4

http://www.bbc.co.uk/schools/gcsebitesiz…


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