# Thread: Help with curve and line

1. ## Help with curve and line

Find the range of values of c for which the line y=c(x-3) does not intersect the curve y+2x=x^2+6

i equate the two eqn together. x² + 6 -2x= cx -3c

and i put b²-4ac<0

so i get (-2-c)² - (24-12c)<0

then i am stuck i dont know how to find the range of values for c

2. Originally Posted by helloying
Find the range of values of c for which the line y=c(x-3) does not intersect the curve y+2x=x^2+6
The function y = c(x - 3) is a straight line, and the function y = x^2 - 2x + 6 is a parabola. You're needing to find values of c so that the straight line passes the parabola off to one side.

To find where the lines do not intersect, we find first where they do intersect; the solution will be the other values of c.

$c(x\, -\, 3)\, =\, x^2\, -\, 2x\, +\, 6$

$cx\, -\, 3c\, =\, x^2\, -\, 2x\, +\, 6$

$0\, =\, x^2\, -\, (c\, +\, 2)x\, +\, 6\, +\, 3c$

Applying the Quadratic Formula, we get:

$x\, =\, \frac{(c\, +\, 2)\, \pm\, \sqrt{(c\, +\, 2)^2\, -\, 4(1)(6\, +\, 3c)}}{2(1)}$

$x\, =\, \frac{(c\, +\, 2)\, \pm\, \sqrt{(c\, +\, 2)^2\, -\, 4(1)(3)(2\, +\, c)}}{2}$

$x\, =\, \frac{(c\, +\, 2)\, \pm\, \sqrt{(c\, +\, 2)^2\, -\, 12(c\, +\, 2)}}{2}$

$x\, =\, \frac{(c\, +\, 2)\, \pm\, \sqrt{(c\, +\, 2)((c\, +\, 2)\, -\, 12)}}{2}$

$x\, =\, \frac{(c\, +\, 2)\, \pm\, \sqrt{(c\, +\, 2)(c\, -\, 10)}}{2}$

So there will be valid solutions (and thus intersections) for (c + 2)(c - 10) > 0. To find the solution, you need to solve the quadratic inequality.

The zeroes of (c + 2)(c - 10) are obviously at c = -2 and c = 10. The parabola corresponding to y = (c + 2)(c - 10) will open upward, so the quadratic will be positive (that is, the parabola will be above the x-axis) on the ends, before the first zero and after the second zero.

So which intervals give intersections? Then which interval does not?