Find all tangent lines to the graph of #y = x^3# that pass through the point. #P(2, 4)#?

Question

Christopher Agresta · Accepted Answer

#{(y-4=(12-6sqrt3)(x-2)),(y-4=3(x-2)),(y-4=(12+6sqrt3)(x-2)):}#

We know two points on the tangent line: the given point #P(2,4)# and the point of tangency located on the graph of #y=x^3#: the point #P(x,x^3)#.

We know that the slope of the tangent line will be equal to the value of the derivative of #y# at the intended value.

The derivative of #y=x^3# is #dy/dx=3x^2#.

Thus, #3x^2# will be equal to the slope of the line passing through the points #P(2,4)# and #P(x,x^3)#.

Setting this up yields:

#3x^2=(x^3-4)/(x-2)#

Now solving:

#3x^2(x-2)=x^3-4#

#3x^3-6x^2=x^3-4#

#2x^3-6x^2+4=0#

To solve this, note that #2-6+4=0#, so #x=1# is a root and #x-1# is a factor of the polynomial.

Do the long division of #(2x^3-6x^2+4)/(x-1)# to find the remaining quadratic factor:

#(x-1)(2x^2-4x-4)=0#

#2(x-1)(x^2-2x-2)=0#

Note that #x^2-2x-2=0# gives solutions of #1+-sqrt3#.

We now know the #x# values for the points of tangency where the tangent lines pass through #P(2,4)#: #1-sqrt3,1,# and #1+sqrt3#.

Find the slopes of the tangent lines by finding the value of the derivative at each of these points:

#f'(1-sqrt3)=3(1-sqrt3)^2=3(1-2sqrt3+3)=12-6sqrt3#

#f'(1)=3(1)^2=3#

#f'(1+sqrt3)=3(1+sqrt3)^2=3(1+2sqrt3+3)=12+6sqrt3#

Then, writing the equations of the three lines with respective slopes of #12-6sqrt3,3,# and #12+6sqrt3# passing through the point #P(2,4)#, we obtain the tangent lines:

#{(y-4=(12-6sqrt3)(x-2)),(y-4=3(x-2)),(y-4=(12+6sqrt3)(x-2)):}#

Charles Anctil · Answer

undefined To find the tangent lines to the graph of y = x^3 that pass through the point P(2, 4), we can use the point-slope form of a line.

First, we differentiate the equation y = x^3 to find the slope of the tangent line at any given point on the graph. The derivative of y = x^3 is dy/dx = 3x^2.

Next, we substitute the x-coordinate of the point P(2, 4) into the derivative to find the slope at that point. dy/dx = 3(2)^2 = 12.

Now, we have the slope of the tangent line at the point P(2, 4). Using the point-slope form of a line, which is y - y1 = m(x - x1), we substitute the values of the point and slope into the equation.

y - 4 = 12(x - 2)

Simplifying the equation, we get:

y - 4 = 12x - 24

Rearranging the equation to the slope-intercept form, we have:

y = 12x - 20

Therefore, the equation of the tangent line to the graph of y = x^3 that passes through the point P(2, 4) is y = 12x - 20.