How do you find the equation of the tangent line to the curve #y=x^4+2x^2-x# at (1,2)?

Question

Luke Alvarez · Accepted Answer

# y = 7x - 5 #

We have; # y = x^4 + 2x^2 - x #

First we differentiate wrt #x#;
# y = x^4 + 2x^2 - x #
# :. dy/dx = 4x^3 + 4x - 1 #

We now find the vale of the derivative at #(1,2)# (and it always worth a quick check to see that #y=2# when #x=1#) we have #dy/dx=4+4-1=7#

So at the tangent passes through the coordinate #(1,2)# and has gradient #m=7#

We now use # y-y_1 = m(x-x_1) # to get the equation of the tangent:

# :. y - 2 = 7 (x - 1) #
# :. y - 2 = 7x - 7 #
# :. y = 7x - 5 #

Luke Alvarez · Answer

The equation of the tangent is #y=7x-5# let #f(x)=x^4+2x^2-x# Then the derivative is #f'(x)=4x^3+4x-1# At the point #(1,2)#, #f'(1)=4+4-1=7# So the slope of the tangent is #m=7# The equation of the line is, #y-2=7(x-1)# #y-7x=-5# graph{(y-x^4-2x^2+x)(y-7x+5)=0 [-5.55, 5.55, -2.773, 2.776]}

Scarlett Allison · Answer

undefined To find the equation of the tangent line to the curve at (1,2), we need to find the slope of the tangent line at that point.

To find the slope, we take the derivative of the function y=x^4+2x^2-x with respect to x.

The derivative of y=x^4+2x^2-x is dy/dx = 4x^3 + 4x - 1.

Substituting x=1 into the derivative, we get dy/dx = 4(1)^3 + 4(1) - 1 = 7.

So, the slope of the tangent line at (1,2) is 7.

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point (1,2) and m is the slope 7, we can substitute the values to find the equation of the tangent line.

Therefore, the equation of the tangent line to the curve y=x^4+2x^2-x at (1,2) is y - 2 = 7(x - 1).