What is the equation of the tangent line of #f(x)=(2x+1)(x+2) # at #x=2#?

Question

What is the equation of the tangent line of #f(x)=(2x+1)(x+2)  # at #x=2#?

Isaiah Allen · Accepted Answer

#y= 13x - 6 #

To make differentiation easier, first make the function more manageable through distribution.

#f(x)=2x^2+5x+2#

Locating the point of tangency is necessary.

#f(2)=2(4)+5(2)+2=20#

The tangent line will pass through the point #(2,20)#.

The power rule allows us to know that

#f'(x)=4x+5#

The slope of the tangent line will be equal to the value of the derivative at #x=2#, or

#f'(2)=4(2)+5=13#

We know the tangent line has a slope of #13# and passes through the point #(2,20)#.

We can write this as an equation in #y=mx+b# form. So far, we know that #m=13#.

#y=13x+b#

Substitute in #(2,20)# for #(x,y)#.

#20=13(2)+b#

#b=-6#

Consequently, the tangent line's equation is

#y=13x-6#

Graphed are #f(x)# and its tangent line:

graph{(y-13x+6)=0 [-4, 6, -10, 50]}((2x+1)(x+2)-y)

Christopher Allers · Answer

undefined To find the equation of the tangent line of f(x)=(2x+1)(x+2) at x=2, we need to find the slope of the tangent line and a point on the line.

First, we find the derivative of f(x) using the product rule: f'(x) = (2x+1)(1) + (x+2)(2) = 4x + 5.

Next, we substitute x=2 into the derivative to find the slope of the tangent line: f'(2) = 4(2) + 5 = 13.

Now, we have the slope of the tangent line, which is 13. To find a point on the line, we substitute x=2 into the original function: f(2) = (2(2)+1)(2+2) = 30.

Therefore, the equation of the tangent line at x=2 is y = 13x + 30.