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

1 Answer
Dec 9, 2016

In slope-intercept form, it is y=84x-507.

Explanation:

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)