Find all of the points on the curve #y = x^3 + 5x^2# where the tangent line is parallel to the line #y = 11x − π#.?

Question

Christopher Allers · Accepted Answer

#(0.8719,4.4639)# and #(-4.2053,14.0549)#

The line #y=11x-pi# has a gradient of 11, thus all lines parallel to this will also have a gradient of 11. So, we must find the points on the curve that will have a tangent with a gradient 11.

To begin we differentiate to find the function that will give us the gradient of the tangent:

#y=x^3+5x^2#

#dy/dx = 3x^2+10x#

As the gradient needs to be 11: we set #dy/dx=11# so:

#3x^2+10x=11#
# -> 3x^2+10x-11=0#

We can now solve the quadratic, but we will need to use the quadratic formula for this:

#x = (-10 +- sqrt(10^2-4(3)(-11)))/(2(3))#

#=(-10+-sqrt(232))/(2(3))#

#=(-10+-2sqrt(58))/(2(3))=-5/3+-sqrt(58)/3#

Thus we have 2 points: #x=-5/3+sqrt(58)/3# and #x=-5/3-sqrt(58)#

Now it is possible to leave it like this or use a calculator to get a rough approximation of each root:

#x~~0.8719# [4 d.p.] and #x~~-4.2053# [4 d.p.]

Substituting these back into the equation we can get the #y# values:

#x~~0.8719 ->y=(0.8719)^3+5(0.8719)^2~~4.4639 # [4 d.p.]
#x~~-4.2053 ->y=(-4.2053)^3+5(-4.2053)^2~~14.0549 # [4 d.p.]

So the points are:

#(0.8719,4.4639)# and #(-4.2053,14.0549)#

To better help with the answer, the plot above shows us the function #y=x^3+5x^2# (blue line), #y=11x-pi# (green line, just above the bottom red line) and the parallel tangents (the two red lines). As can be seen, the two red lines run parallel to the green line and intersect the blue line at the points calculated.

Aurora Alvarado · Answer

undefined To find the points on the curve $ y = x^3 + 5x^2 $ where the tangent line is parallel to the line $ y = 11x - \pi $, we first need to find the derivative of the curve. The derivative of $ y = x^3 + 5x^2 $ is $ y' = 3x^2 + 10x $.

Since the tangent line to the curve at a given point has the same slope as the derivative at that point, we set $ y' $ equal to the slope of the given line, which is $ 11 $. So, we have the equation $ 3x^2 + 10x = 11 $.

Solving this equation will give us the x-coordinates of the points where the tangent line is parallel to the given line. After finding the x-coordinates, we can substitute them back into the original equation $ y = x^3 + 5x^2 $ to find the corresponding y-coordinates.

Solving $ 3x^2 + 10x = 11 $ gives us $ x = -1 $ and $ x = 1 $.

Substituting $ x = -1 $ into $ y = x^3 + 5x^2 $, we get $ y = (-1)^3 + 5(-1)^2 = -4 $.

Substituting $ x = 1 $ into $ y = x^3 + 5x^2 $, we get $ y = 1^3 + 5(1)^2 = 6 $.

So, the points on the curve $ y = x^3 + 5x^2 $ where the tangent line is parallel to the line $ y = 11x - \pi $ are $ (-1, -4) $ and $ (1, 6) $.