a. Write the equation of the line that represents the linear approximation to the following function at the given point a. b. Use the linear approximation to estimate the given quantity. c. Compute the percent error in the approximation, $100 \cdot \frac{\text { |approximation - exact| }}{\text { |exact| }}$, where the exact value is given by a calculator. \[ f(x)=1-x^{2} \text { at } a=2 ; f(1.9) \] a. $L(x)=-4 x+5$ b. Using the linear approximation, $\mathrm{f}(1.9) \approx-2.6$. (Type an integer or a decimal.) c. The percent error in the approximation is $30 \%$. (Round to three decimal places as needed.)

Step 1 ：Find the derivative of the function \(f(x) = 1 - x^2\), which is \(f'(x) = -2x\).

Step 2 ：Substitute \(a = 2\) into \(f'(x)\) and \(f(x)\) to get \(f'(2) = -4\) and \(f(2) = -3\).

Step 3 ：Use the formula for the tangent line to find the equation of the line, which is \(L(x) = -4(x - 2) - 3 = -4x + 5\).

Step 4 ：Substitute \(x = 1.9\) into the equation of the line to get \(L(1.9) = -4(1.9) + 5 = -2.6\).

Step 5 ：Find the exact value of \(f(1.9)\) using a calculator, which is \(f(1.9) = 1 - (1.9)^2 = -2.61\).

Step 6 ：Substitute these values into the formula for the percent error to get \(100 \cdot \frac{|-2.6 - (-2.61)|}{|-2.61|} = 100 \cdot \frac{0.01}{2.61} \approx 0.383 \%\).

Step 7 ：\(\boxed{0.383 \%}\) is the percent error in the approximation.