The gradient to the curve y=2x^2+x-1 at point A is the perpendicular to the straight line 5y=-x. find coordinate of A.

Question

anonymous · Answer

gradient in this case means slope, right ?
find the slope of the straight line first,
$$5y = -x$$$$y = -\frac{1}{5}x$$$$m_{1} = -1/5$$because the slope at point A is perpendicular to the slope of the straight line,$$m_{1}m_{2} = -1$$$$m_{2} = 5$$then we can find point A by using the first derivative of the curve$$y = 2x^2 + x - 1$$$$y' = 4x + 1$$$$5 = 4x + 1$$$$x = 1$$$$y = 2(1)^2 + 1 - 1$$$$y = 2$$point A is (1,2)

anonymous · Answer

How to know when to use derivative??

anonymous · Answer

y′=4x+1
5=4x+1<< i dont understand this part?? where do u get the five.
x=1

anonymous · Answer

to find the slope of a curve at a certain point you need to use the first derivative of the curve

anonymous · Answer

$$y' = 4x + 1$$$$m_{2} = 4x + 1$$

anonymous · Answer

TQ. Very much!!
From that, how to find the equation of the tangent  to the curve at point A?

anonymous · Answer

recall the equation of a line$$y - y_{1} = m(x - x_{1})$$plug (1,2) for x1 and y1, and 5 for m

anonymous · Answer

Is it y=5x-3 ? 
TQ very much.