Question

1. at what point on the graph of $f(x)=x^{2}-4$ is the tangent line parallel to the line $6x - 3y = 2$?

(a) $(1,-3)$ (b) $(1,2)$ (c) $(2,0)$ (d) $(2,4)$

Explanation:

Step1: Find the slope of the given line

Rewrite \(6x - 3y=2\) as \(y = 2x-\frac{2}{3}\). The slope \(m = 2\).

Step2: Find the derivative of \(f(x)\)

Using the power rule, if \(f(x)=x^{2}-4\), then \(f^{\prime}(x)=2x\).

Step3: Set the derivative equal to the slope of the line

Set \(2x = 2\), solving for \(x\) gives \(x = 1\).

Step4: Find the \(y\) - value

Substitute \(x = 1\) into \(f(x)=x^{2}-4\), \(f(1)=1^{2}-4=-3\).

Answer:

A. \((1,-3)\)