Answer:
To find the equation of the tangent line to the graph of the function \(f(x) = x + 3x - 3\) at the point \(x = -4\), we need to follow these steps:
1. Find the value of \(f(-4)\) at \(x = -4\), which will give us the y-coordinate of the point on the graph.
2. Find the derivative of the function \(f(x)\) to get the slope of the tangent line at \(x = -4\).
3. Use the point-slope form of the equation of a line to write the equation of the tangent line.
Let's go through each step:
1. Find \(f(-4)\):
\(f(-4) = -4 + 3(-4) - 3 = -4 - 12 - 3 = -19\)
2. Find the derivative \(f'(x)\) of \(f(x)\):
\(f(x) = x + 3x - 3\)
\(f'(x) = \frac{d}{dx}(x) + \frac{d}{dx}(3x) - \frac{d}{dx}(3)\)
\(f'(x) = 1 + 3 - 0\)
\(f'(x) = 4\)
3. Use the point-slope form of the equation of a line to write the equation of the tangent line:
The point-slope form is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is a point on the line.
We have \(m = 4\) (the derivative at \(x = -4\)) and \((x_1, y_1) = (-4, -19)\) (the point on the graph).
Plug these values into the point-slope form:
\(y - (-19) = 4(x - (-4))\)
Simplify:
\(y + 19 = 4(x + 4)\)
Distribute the 4 on the right side:
\(y + 19 = 4x + 16\)
Now, isolate \(y\):
\(y = 4x + 16 - 19\)
\(y = 4x - 3\)
So, the equation of the tangent line to the graph of the function \(f(x) = x + 3x - 3\) at the point \(x = -4\) is:
\[y = 4x - 3\]
This is in the desired format \(y = mx + b\), where \(m\) is the slope (4) and \(b\) is the y-intercept (-3).
Explanation: