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Need a graph based solution please

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by [email protected] » Sun Aug 18, 2013 6:14 pm
The equation x = 2y2 + 5y - 17, describes a parabola in the xy coordinate plane. If line l, with slope of 3, intersects the parabola in the upper-left quadrant at x = -5, the equation for l is
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by [email protected] » Sun Aug 18, 2013 9:37 pm
Hi shibsriz,

What is the source of this question? And what are the 5 answer choices?

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by vipulgoyal » Wed Aug 21, 2013 2:56 am
is it

y = 3/4x + 15 ??
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by ganeshrkamath » Wed Aug 21, 2013 3:32 am
[email protected] wrote:The equation x = 2y2 + 5y - 17, describes a parabola in the xy coordinate plane. If line l, with slope of 3, intersects the parabola in the upper-left quadrant at x = -5, the equation for l is
Let the line l be defined as:
y = mx + c
Here slope = m = 3
y = 3x + c ____________ (1)

Since this intersects the parabola at x = -5
-5 = 2y^2 + 5y - 17
2y^2 + 5y - 12 = 0
y = (-5 +/- sqrt(25 + 96))/4
y = (-5 +/- sqrt(121))/4
y = (-5 +/- 11)/4
y = -4 or y = 3/2
Since the point of intersection lies on the upper left quadrant, y > 0.
So y = 3/2

Substitute the values of x and y in
3/2 = 3(-5) + c
3/2 = -15 + c
c = 33/2

So the equation of the line is:
[spoiler]y = 3x + (33/2)[/spoiler]

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