00:01
For what point on the curve, y equals 5x squared plus 6x is the slope of a tangent line equal to 46? so we know that the slope of a tangent line to this curve or to any curve is equal to the derivative at that point.
00:25
So slope of the tangent line to the curve at...
00:49
Xy on the curve is the derivative of y respect to x so the derivative of y respect to x in this case is 10x plus 6 and then we want to find x for which because remember the point is on the curve it means it has a first coordinate x and the second coordinate y which is given by the this expression in terms of x.
01:45
But the derivative of y respect to x is evaluated at x, at the corresponding x value.
01:54
That is the first coordinate of the point.
01:58
So we want to find that first coordinate x for which derivative is equal to 46.
02:09
That is 10x plus 6, which is the derivative of y respect to x equal 46.
02:16
From here we get 10 x equal 40 which is 46 minus 6 and so x is 40 over 10 that is x equal 4.
02:31
But that's the first coordinate.
02:36
Second coordinate of the point which is the point of tangency is y equal 5 times 4...