Francisco E. answered • 05/16/14
Tutor
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Francisco; Civil Engineering, Math., Science, Spanish, Computers.
The hyperbola opens horizontally, the form of the equation is ((x-0)^2)/1- ((y-3)^2)/16 = 1, we may write it in a different way: ((X^2)/2)-(((y^2)/16)-(6y/16)+(9/16))=1; making a common denominator=> (8x^2 -y^2+6y-9)/16=1=> 8x^2 - y^2 + 6y -25=0
Then we may define A=8; C=-1; D=0; E=6; F=-25. Now we are ready to graph the hyperbola as follows:
h=y(center)=-D/2A=0; k=x(center=-E/2C=-6/(2*-1)= 3; a= distance from center to vertices= √(-*-1) =1, the hyperbola opens horizontally , center located at point (0,3), Vertices at (-1,3) and (1,3)
Now you can graph it!!!!!!!!!!!!!!!!!!!!!!
