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Mathematics 16 Online
OpenStudy (anonymous):

How do you write the equation in standard form, then determine the vertex, axis and direction in which each parabola opens? x^2 + 2x - 3y + 19 = 0

OpenStudy (anonymous):

write the eq in general form \[X^2=4aY\]

OpenStudy (anonymous):

vertex is X=0 Y=0

OpenStudy (anonymous):

\[x^2+2x+1=3y-19+1\]

OpenStudy (anonymous):

Standard form is y = ax^2 + bx + c So add 3y to both sides x^2 + 2x + 19 = 3y now divide by 3 y = 1/3 x^2 + 2/3 x + 19/3 It will open up along the y-axis. The vertex is (-1, 6)

OpenStudy (anonymous):

\[(x+1)^2=3y+18\]

OpenStudy (anonymous):

\[(x+1)^2=3(y+6)\]

OpenStudy (anonymous):

so \[X^2=3Y\]

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