The problem in the image asks to find the slope of the tangent line to the curve $y = x^2 - 3x + 2$ at two points: $x = -2$ and $x = 1.5$.

Here’s how we can solve this:

Step 1: Find the derivative of the curve

To find the slope of the tangent line, we need to compute the derivative of the function $y = x^2 - 3x + 2$. The derivative of this function gives the slope of the tangent line at any point $x$.

$\frac{dy}{dx} = \frac{d}{dx}(x^2 - 3x + 2) = 2x - 3$

Step 2: Evaluate the derivative at the given points

1. For $x = -2$: $\frac{dy}{dx}\bigg|_{x=-2} = 2(-2) - 3 = -4 - 3 = -7$ So, the slope at $x = -2$ is $-7$.

2. For $x = 1.5$: $\frac{dy}{dx}\bigg|_{x=1.5} = 2(1.5) - 3 = 3 - 3 = 0$ So, the slope at $x = 1.5$ is $0$.

Final Answer:

• The slope of the tangent line at $x = -2$ is $-7$.
• The slope of the tangent line at $x = 1.5$ is $0$.

Would you like more details or further explanation on any of the steps?

Here are some related questions for further practice:

1. How do you find the equation of a tangent line given a curve and a point?
2. What is the geometrical interpretation of a slope of zero for a tangent line?
3. How would you compute the slope of the tangent line for higher-order polynomials?
4. How can you use derivatives to determine if a function is increasing or decreasing at a specific point?
5. Can you find the second derivative of the function and explain its significance?

Tip: The slope of the tangent line tells you the rate of change of the function at a specific point. When the slope is zero, the tangent line is horizontal, indicating a local maximum or minimum.