Find to find equation of the tangent line y=x^4-x^3-x^2-x+1 at the point x=1
Is the calculus?
If so, take the derivative of y which would be 4x^3 - 3x^2-2x -1
Let x =1 to find the slope
4-3-2-1 = m = -2
find the value of y when x = 1
You will then have the slope and a point to use to find the equation using y=mx+b
so i found the y=-1
then i subbed into y=mx+b and got -1=-2(1)+b
1=b
then my equation would be -1=-2(1)+1
is that right?
To find the equation of the tangent line to the curve y = x^4 - x^3 - x^2 - x + 1 at the point x = 1, we need to find two key pieces of information: the slope of the tangent line at that point and the coordinates of the point.
1. Find the slope:
The slope of the tangent line at a given point can be found by taking the derivative of the function and evaluating it at that point.
So, let's find the derivative of the function y = x^4 - x^3 - x^2 - x + 1:
dy/dx = 4x^3 - 3x^2 - 2x - 1
Evaluate the derivative at x = 1:
dy/dx(x = 1) = 4(1)^3 - 3(1)^2 - 2(1) - 1 = 4 - 3 - 2 - 1 = -2
So, the slope of the tangent line at x = 1 is -2.
2. Find the coordinates of the point:
To find the y-coordinate at x = 1, substitute the value of x into the original function:
y = (1)^4 - (1)^3 - (1)^2 - (1) + 1 = 1 - 1 - 1 - 1 + 1 = -1
Therefore, the coordinates of the point on the curve at x = 1 are (1, -1).
Now that we have the slope (-2) and a point on the line (1, -1), we can use the point-slope form of a line to find the equation of the tangent line.
Point-slope form: y - y1 = m(x - x1)
Substituting the values we found:
y - (-1) = -2(x - 1)
y + 1 = -2x + 2
Simplifying the equation:
y = -2x + 1
Therefore, the equation of the tangent line to the curve y = x^4 - x^3 - x^2 - x + 1 at the point x = 1 is y = -2x + 1.