For #f(x) =3 x^2 + 6 x#, what is the equation of the line tangent to #x =13 #?

Step 1: Find the first derivative #f'(x)# using the power rule:

#d/dxax^n=nax^(n-1)#

So #f(x)=3x^2+6x#

#=>f'(x)=6x+6#

This gives us an equation to find the slope of #f(x)# at any given #x#-value.

Step 2: Find the slope at #x=13# by plugging this #x#-value into #f'(x).#

#f'(x)=6x+6#
#=>f'(13)=6(13)+6#
#color(white)(=>f'(13))=78+6#
#color(white)(=>f'(13))=84#

So when #x=13#, the slope is #m=f'(13)=84#.

Step 3: Find the #y#-coordinate for #x=13# using #y=f(x)#.

#f(x)=3x^2+6x#
#=>f(13)=3(13)^2+6(13)#
#color(white)(=>f(13))=3(169)+78#
#color(white)(=>f(13))=507+78#
#color(white)(=>f(13))=585#

So when #x=13#, we have #y=f(13)=585#.

Step 4: Plug these known values for #x#, #y#, and #m# into your favourite line equation. (I will use slope-point form: #y=m(x-x_1)+y_1#.)

#y=m(x-x_1)+y_1#
#=>y=84(x-13)+585#
#=>y=84x-1092+585#
#=>y=84x-507#

This is our equation for the line tangent to #f(x)# at #x=13#.

Note:

For consistency with calculus, the line equation may be written

#y=f'(x_0)(x-x_0)+f(x_0)#

The calculations are identical; the only difference is the symbols that are used as placeholders for the information we need:

#m=f'(x_0)#
#x_1=x_0#
#y_1=f(x_0)#