Friday, January 8, 2010

How to find the gradient from this equation?

How to find the value(s) of m (gradient) when the line y=mx-4 is a tangent to parabola y= x^2-2x+12?How to find the gradient from this equation?
You can find the general gradient of the parabola by differentiating.


If y = x^2 - 2x + 12


dy/dx = 2x-2





dy/dx is the gradient so it = m





You can substitute this into the second equation to get


y = (2x-2)x - 4


= 2x^2 - 2x - 4





If you put this equal to 0, this will show you the x-coordinates of the equation





0 = 2x^2 - 2x - 4


0 = (x+1)(x-2) so: x = -1 or +2





From there, we can substitute these values to find the two values of y using the equation of the parabola


(1) y = (-1)^2 -2(-1) + 12 = 15


(2) y = (2)^2 - 2(2) + 12 = 12





So the two coordinates where the line is a tangent are


(-1, 15) and (2,12)





We can substitute this into the linear equation then as:


15 = -m -4 so m = -19


and 12 = 2m - 4 so m = 8











Hope that all helps and makes senseHow to find the gradient from this equation?
There once was this facebook group called ';I Wish I Were Your Derivative So I Could Lie Tangent To Your Curves';.





A derivative is defined as the limit as h approaches 0 of the equation:





(f(x + h) - f(x) ) / (h), which is basically like saying ';rise over run'; when run is effectively 0. So you begin the process of plugging in. (x + h) ^ 2 - 2(x + h) + 12 - (x^2 - 2x + 12) = (2xh + h^2 - 2h) is your numerator, and divide it all by h, and you get 2x + h - 2, and since h approaches 0, you have 2x - 2 is the formula for the slope of the function at point x. I won't spoil the rest by fully solving it for you (since now the hard part is out of the way), and I make no guarantees I didn't go the long way of solving this, but hopefully you learned something from this computer science nerd.

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