Find the equation of the tangent line at the point (0,1) of the graph of the function f(x) = x^3 - 2x + 1 ?​

9514 1404 393

  y = -2x +1

Step-by-step explanation:

The derivative of the function is ...

  f'(x) = 3x^2 -2

so the slope at x=0 is f'(0) = -2. In slope-intercept form, the equation of the tangent line is ...

  y = -2x +1

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

answer: 5

step-by-step explanation: