Posted by **jan** on Monday, November 28, 2011 at 11:51am.

graph the parabola y=2x^2-6x-3. plot point on the parabola A[1,-7] and draw a line through A with an angle of inclination equal to 30 degrees. then find the equation of the line and its second point of intersection B, with the parabola

- precalculus -
**Steve**, Monday, November 28, 2011 at 12:56pm
Just take this one step at a time:

a line with inclination 30° has slope tan30° = 1/√3

Now we have a point and a slope, so that line is

(y+7)/(x-1) = 1/√3

y = 1/√3 x - (7+1/√3)

Now find where that line intersects the parabola:

2x^2 - 6x - 3 = 1/√3 x - (7 + 1/√3)

x = 1

x = 2 + 1/(2√3) or 2.288

The second point of intersection is thus

(2 + 1/(2√3) , 1/(2√3) - 41/6)

or (2.288,-6.256)

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