For f(x) =3 x^2 + 6 x, what is the equation of the line tangent to x =13 ?
1 Answer
In slope-intercept form, it is
Explanation:
Step 1: Find the first derivative
d/dxax^n=nax^(n-1)
So
=>f'(x)=6x+6
This gives us an equation to find the slope of
Step 2: Find the slope at
f'(x)=6x+6
=>f'(13)=6(13)+6
color(white)(=>f'(13))=78+6
color(white)(=>f'(13))=84
So when
Step 3: Find the
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
Step 4: Plug these known values for
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
Note:
For consistency with calculus, the line equation may be written
The calculations are identical; the only difference is the symbols that are used as placeholders for the information we need: