1. x^{2} - x - 12 = x +c
x^{2} - 2x - 12 = c
What value would I substitute for x? I think it has something to do with the equations being tangental, but I can't remember exactly.
2. I know that I have to substitute coordinates into the equation ax^{2} + bx + c and solve simultaneously, but I don't know how to solve simultaneous equations with three variables.
Now we have the curve y = x^2 - x -12. Slope of tangent to the curve is given by dy/dx. for this curve dy/dx = 2x -1 Let the point of contact be (h,k)
Thus slope of tangent = 2h-1. The slope of the given line y = x + c is 1. Hence 2h - 1 = 1 that gives h = 1. Form the equation of the curve , by pugging in value of h=1 for x we get ythe value of k since (h,k) lies on the curve.
That gives k = 12. Hence the point of contact is ( 1, - 12 ) and slope of tangent is = 1. Thus its equation is y + 12 = 1 ( x - 1 ) that is y = x - 13. Comparing it with the given equation we get c = -13.
He's using calculus, but we may use a pre-calculus technique instead, since you posted in the Algebra forum. You correctly found:
$\displaystyle x^2-2x-12=c$
If we arrange this in standard quadratic form, we get:
$\displaystyle x^2-2x-(12+c)=0$
Now, if the line and parabola are tangent, this means they share only one point in common, i.e., the above quadratic has only one root. What condition on a quadratic must hold in order for there to be only one root?
Alright, thanks. I'll have a go solving it tomorrow -- it's getting late here.
This may sound ridiculously stupid, but why does the equation become x^2 - 2x - (12 + c) = 0 as opposed to something like, say, x^2 - 2x + c -12 = 0? Specifically, why does the c get tied to the 12 when it's moved?
Since c is a constant, it should be grouped with the constant 12, as the other two terms have a power of x as factors. Since it began with a positive sign on the right, when we move it to the left, we are essentially subtracting c from both sides.
This is the solution to part ii of problem 1:
Since the tangent line and the given curve have a point in common we have:
$\displaystyle mx + 6 = -2x^2 - 6x + 2$
$\displaystyle 2x^2 + (6 + m)x + 4 = 0$
slope of $\displaystyle m = \displaystyle\frac{dy}{dx}$ where $\displaystyle \displaystyle\frac{dy}{dx} = \displaystyle\frac{d}{dx} (-2x^2 - 6x + 2) = -4x - 6 = 0$
so $\displaystyle 6 + m = 6 - 4x - 6 = -4x$
$\displaystyle 2x^2 -4x^2 + 4 = 0$
$\displaystyle -2x^2 + 4 = 0$; $\displaystyle x^2 = 2$; $\displaystyle x = \pm \sqrt{2}$
if $\displaystyle x = \sqrt{2}$; $\displaystyle m = -4\sqrt{2} - 6 \approx -11.657 $
if $\displaystyle x = -\sqrt{2}$; $\displaystyle m = 4\sqrt{2} - 6 \approx -.343 $
Since this question is posted in the Algebra forum, we should use pre-calculus methods if we are going to truly help the OP. I was hoping the OP would be able to apply the method of equating the discriminant to zero, as we did in the first part of the first problem, to solve the second part.