Question

Write the equation of the tangent line to function
f(x)
at
x = −3
if
f(−3) = 6
and
f '(−3) =
4
5
.

Answer 1

To find the equation of the tangent line to the function \( f(x) \) at \( x = -3 \), we can use the point-slope form of the equation of a line, which is given by:

\[
y - y_1 = m (x - x_1)
\]

where \( m \) is the slope of the tangent line, \( (x_1, y_1) \) is a point on the line, and \( y \) and \( x \) are the variables.

In this case, we have:

- \( x_1 = -3 \)
- \( y_1 = f(-3) = 6 \)
- The slope \( m = f'(-3) = \frac{4}{5} \)

Now we can substitute these values into the point-slope form:

\[
y - 6 = \frac{4}{5} (x - (-3))
\]

This simplifies to:

\[
y - 6 = \frac{4}{5} (x + 3)
\]

Next, we can distribute \( \frac{4}{5} \):

\[
y - 6 = \frac{4}{5}x + \frac{4}{5} \cdot 3
\]
\[
y - 6 = \frac{4}{5}x + \frac{12}{5}
\]

Now, add 6 (or \( \frac{30}{5} \)) to both sides to isolate \( y \):

\[
y = \frac{4}{5}x + \frac{12}{5} + \frac{30}{5}
\]
\[
y = \frac{4}{5}x + \frac{42}{5}
\]

Thus, the equation of the tangent line at \( x = -3 \) is:

\[
y = \frac{4}{5}x + \frac{42}{5}
\]