precalculus
posted by jan .
graph the parabola y=2x^26x3. 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

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)/(x1) = 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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