How do you find the equation of the line tangent to #y = (1+2x)^2#, at (1,9)?

Question

William Abdalla · Accepted Answer

#y=12x-3#

We got: #y=(2x+1)^2=4x^2+4x+1#.

First, we find the slope of the tangent line at #(1,9)#, which is the derivative of #y# at #x=1#.

#:.y'=8x+4#

At #x=1#,

#m=8*1+4#

#=8+4#

#=12#

So, the slope of the tangent line is #12#.

Now, we use the point-slope form, which is:

#y-y_0=m(x-x_0)#

where #(x_0,y_0)# are the original coordinates of the original function, #y#.

And, we get,

#y-9=12(x-1)#

#y-9=12x-12#

#y=12x-12+9#

#y=12x-3#

A graph shows it, but it's not really clear:

Aurora Alvarado · Answer

undefined To find the equation of the line tangent to the curve y = (1+2x)^2 at the point (1,9), we need to find the slope of the tangent line and the coordinates of the point of tangency.

First, we find the derivative of the given function y = (1+2x)^2 with respect to x.

Using the chain rule, the derivative is: dy/dx = 2(1+2x)(2) = 4(1+2x).

Next, we substitute x = 1 into the derivative to find the slope of the tangent line at x = 1: dy/dx = 4(1+2(1)) = 12.

Therefore, the slope of the tangent line is 12.

Now, we have the slope of the tangent line and the point of tangency (1,9). We can use the point-slope form of a linear equation to find the equation of the tangent line.

Using the point-slope form: y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point of tangency, we substitute the values:

y - 9 = 12(x - 1).

Simplifying the equation, we get the equation of the tangent line:

y - 9 = 12x - 12.

Rearranging the equation, we have:

y = 12x - 3.

Therefore, the equation of the line tangent to y = (1+2x)^2 at (1,9) is y = 12x - 3.