Find the equation of the tangent line to the graph of f(x)=[2x^2]-1 at (-1,1)?

Answer 1

f'(x)= 4x

So the slope of the line is 4(-1), which is -4

So the equation we get is:

y-1=-4(x+1)

Answer 2

the derivative of 2x^2 is 4x and it give the slope at any point on the graph of 2x^2.

At (-1,1) the slope is 4(-1) = -4

Use point slope equation of line: y-y1 = m (x-x1) at point (-1,1) yields:

y - 1 = -4 (x- -1)
y-1 = -4x -4
y=-4x-3 (answer)

Answer 3

f ` (x) = 4x
f ` (-1) = - 4 is gradient , m
y - b = m (x - a)
y - 1 = (- 4).(x + 1)
y = - 4x - 3 is equation of tangent thro` (-1 , 1)

Answer 4

y'=4x

m=4.(-1)=-4
y-1=-4(x+1)
y=-4x-4+1

tangen line y=-4x-3