How do you find the tangents to the curve #y= x^3 + x# at the points where the slope is 4?

1 Answer
mason m
Apr 17, 2016

#y=4x+2# and #y=4x-2#

Explanation:

When the tangent line to a function has a slope of #4#, this is the same as saying that the derivative of the function is equal to #4# at that same point.

Thus, we have to find the derivative of the function and set it equal to #4#.

To differentiate #y=x^3+x#, we will use the power rule. Through the power rule, we see that #dy/dx=3x^2+1#. So, we then set #dy/dx=4# since this is the slope of the tangent line at the desired points. This gives us the equation

#3x^2+1=4" "=>" "3x^2=3" "=>" "x^2=1#

Which implies that #x=+-1#. Thus, we have two points. Determine their coordinates by plugging #-1,1# back into the equation for #y#.

#y(-1)=(-1)^3+(-1)=-2" "" "(-1,-2)#

#y(1)=1^3+1=2" "" "" "" "" "" "" "" "(1,2)#

If you wish to determine the equations of the tangent lines, plug the points and slope of #4# into equations in point-slope form, which takes the point #(x_1,y_1)# and slope #m# and relates them as the equation:

#y-y_1=m(x-x_1)#

Thus, we have the two equations

#mathbf(1")")" "y-(-2)=4(x-(-1))#
#" "" "y+2=4(x+1)#
#" "" "y=4x+2#

#mathbf(2")")" "y-2=4(x-1)#
#" "" "y=4x-2#

Related questions
See all questions in Tangent Line to a Curve
Impact of this question
15298 views around the world
You can reuse this answer
Creative Commons License