Questions: Find the slope of the tangent to the graph of (x^2-4)/(x^2+x+1) at the point (-2,0). Write your answer as reduced fraction. Slope of the tangent line =

$Find the slope of the tangent to the graph of (x^2-4)/(x^2+x+1) at the point (-2,0). Write your answer as reduced fraction. Slope of the tangent line =$

Transcript text: Find the slope of the tangent to the graph of $\frac{x^{2}-4}{x^{2}+x+1}$ at the point $(-2,0)$. Write your answer as reduced fraction. Slope of the tangent line $=$ $\square$ Submit Question

Solution

Solution Steps

To find the slope of the tangent to the graph of the function at a given point, we need to compute the derivative of the function and evaluate it at the specified point. The derivative will give us the slope of the tangent line at any point on the curve.

Differentiate the function $\frac{x^2 - 4}{x^2 + x + 1}$ using the quotient rule.
Evaluate the derivative at the point $x = -2$.

Step 1: Differentiate the Function

To find the slope of the tangent line to the graph of the function $ f(x) = \frac{x^2 - 4}{x^2 + x + 1} $ at a given point, we first need to compute the derivative of the function. Using the quotient rule, the derivative $ f'(x) $ is given by:

\[ f'(x) = \frac{(2x)(x^2 + x + 1) - (x^2 - 4)(2x + 1)}{(x^2 + x + 1)^2} \]

Simplifying, we have:

\[ f'(x) = \frac{2x(x^2 + x + 1) - (2x + 1)(x^2 - 4)}{(x^2 + x + 1)^2} \]

Step 2: Evaluate the Derivative at the Given Point

Next, we evaluate the derivative at the point $ x = -2 $ to find the slope of the tangent line at that point:

\[ f'(-2) = \frac{2(-2)((-2)^2 + (-2) + 1) - ((-2)^2 - 4)(2(-2) + 1)}{((-2)^2 + (-2) + 1)^2} \]

Simplifying further, we find:

\[ f'(-2) = \frac{-4(4 - 2 + 1) - (4 - 4)(-4 + 1)}{(4 - 2 + 1)^2} = \frac{-4 \times 3}{3^2} = \frac{-12}{9} = -\frac{4}{3} \]