Answer:
To find the equation of the tangent line to the graph of the function \(f(x) = (x^2 + 6)(x - 4)\) at the point \((1, -21)\), you can follow these steps:
1. Find the derivative of \(f(x)\) to get the slope of the tangent line.
2. Use the point-slope form of the equation of a line to find the equation of the tangent line.
Let's start with step 1:
\(f(x) = (x^2 + 6)(x - 4)\)
Use the product rule to find the derivative \(f'(x)\):
\(f'(x) = (x - 4) \cdot \frac{d}{dx}(x^2 + 6) + (x^2 + 6) \cdot \frac{d}{dx}(x - 4)\)
\(f'(x) = (x - 4) \cdot (2x) + (x^2 + 6) \cdot 1\)
\(f'(x) = 2x(x - 4) + (x^2 + 6)\)
Now, let's find the slope of the tangent line at the point \((1, -21)\) by plugging \(x = 1\) into \(f'(x)\):
\(f'(1) = 2(1)(1 - 4) + (1^2 + 6)\)
\(f'(1) = 2(-3) + 7\)
\(f'(1) = -6 + 7\)
\(f'(1) = 1\)
So, the slope of the tangent line at \((1, -21)\) is \(m = 1\).
Now, move on to step 2, where we use the point-slope form of the equation of a line:
\(y - y_1 = m(x - x_1)\)
where \((x_1, y_1)\) is the point \((1, -21)\) and \(m\) is the slope we just found.
Plugging in the values:
\(y - (-21) = 1(x - 1)\)
Simplify:
\(y + 21 = x - 1\)
Now, isolate \(y\):
\(y = x - 1 - 21\)
\(y = x - 22\)
So, the equation of the tangent line to the graph of \(f(x) = (x^2 + 6)(x - 4)\) at the point \((1, -21)\) is \(y = x - 22\).
Explanation: