Questions: The graphs of (f(x)=x^2-2 x+2) and (g(x)=5 x+20) intersect at the point ((-2,10)). The (acute) angle of intersection (in radians) of these graphs at ((-2,10)) is:

Transcript text: The graphs of $f(x)=x^{2}-2 x+2$ and $g(x)=5 x+20$ intersect at the point $(-2,10)$. The (acute) angle of intersection (in radians) of these graphs at $(-2,10)$ is:

Solution

Solution Steps

Step 1: Find the derivatives of $ f(x) $ and $ g(x) $

To determine the angle of intersection, we first need the slopes of the tangent lines to the graphs of $ f(x) $ and $ g(x) $ at the point $(-2, 10)$.

The derivative of $ f(x) = x^{2} - 2x + 2 $ is: \[ f'(x) = 2x - 2 \] At $ x = -2 $: \[ f'(-2) = 2(-2) - 2 = -4 - 2 = -6 \]
The derivative of $ g(x) = 5x + 20 $ is: \[ g'(x) = 5 \] At $ x = -2 $: \[ g'(-2) = 5 \]

Step 2: Calculate the angle between the tangent lines

The angle $ \theta $ between two lines with slopes $ m_1 $ and $ m_2 $ is given by: \[ \tan(\theta) = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \] Here, $ m_1 = -6 $ (slope of $ f(x) $) and $ m_2 = 5 $ (slope of $ g(x) $).

Substitute the values: \[ \tan(\theta) = \left| \frac{5 - (-6)}{1 + (-6)(5)} \right| = \left| \frac{11}{1 - 30} \right| = \left| \frac{11}{-29} \right| = \frac{11}{29} \]

Now, calculate $ \theta $: \[ \theta = \arctan\left(\frac{11}{29}\right) \]

Step 3: Compute the numerical value of $ \theta $

Using a calculator: \[ \theta = \arctan\left(\frac{11}{29}\right) \approx 0.3614 \text{ radians} \]