Question

find an equation for the tangent line to the graph of the given function at (2, -2). f(x) = x² - 6 find an equation for the tangent line to the graph of f(x) = x² - 6 at (2, -2). y = (type an expression using x as the variable.)

Explanation:

Step1: Find the derivative of \( f(x) \)

The function is \( f(x) = x^2 - 6 \). The derivative \( f'(x) \) using the power rule (\( \frac{d}{dx}x^n = nx^{n - 1} \)) is \( f'(x) = 2x \).

Step2: Find the slope at \( x = 2 \)

Substitute \( x = 2 \) into \( f'(x) \). So \( f'(2) = 2\times2 = 4 \). The slope \( m \) of the tangent line is 4.

Step3: Use point - slope form

The point - slope form of a line is \( y - y_1 = m(x - x_1) \), where \( (x_1,y_1)=(2,-2) \) and \( m = 4 \).
Substitute the values: \( y - (-2)=4(x - 2) \).
Simplify: \( y + 2 = 4x-8 \).
Then \( y=4x - 8 - 2 \), so \( y = 4x-10 \).

Answer:

\( 4x - 10 \)