Find the slope of the secant lines over the specified intervals for the function 𝑓(𝑥) = − ଵ
ଶ 𝑥ଶ + 2𝑥. ) [2, 3]
b) [2, 2.1]
b) Estimate the slope of the tangent line at 𝑥 = 2
d) Use your estimate to find the equation of the tangent line at 𝑥 = 2

We have the function:

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

### a) Find the slope of the secant line over the interval \([2, 3]\):

The slope of the secant line over an interval \([x_1, x_2]\) is given by:

\[
m_{\text{sec}} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}
\]

In this case, \(x_1 = 2\) and \(x_2 = 3\).

First, calculate \(f(2)\) and \(f(3)\):

\[
f(2) = -\frac{1}{2}(2^2) + 2(2) = -\frac{1}{2}(4) + 4 = -2 + 4 = 2
\]
\[
f(3) = -\frac{1}{2}(3^2) + 2(3) = -\frac{1}{2}(9) + 6 = -4.5 + 6 = 1.5
\]

Now, find the slope:

\[
m_{\text{sec}} = \frac{1.5 - 2}{3 - 2} = \frac{-0.5}{1} = -0.5
\]

So, the slope of the secant line over \([2, 3]\) is \(-0.5\).

---

### b) Find the slope of the secant line over the interval \([2, 2.1]\):

Now, we do the same procedure for the interval \([2, 2.1]\).

Calculate \(f(2.1)\):

\[
f(2.1) = -\frac{1}{2}(2.1^2) + 2(2.1) = -\frac{1}{2}(4.41) + 4.2 = -2.205 + 4.2 = 1.995
\]

Now, find the slope:

\[
m_{\text{sec}} = \frac{1.995 - 2}{2.1 - 2} = \frac{-0.005}{0.1} = -0.05
\]

So, the slope of the secant line over \([2, 2.1]\) is approximately \(-0.05\).

---

### c) Estimate the slope of the tangent line at \(x = 2\):

The slope of the tangent line at \(x = 2\) is the limit of the secant slopes as the interval gets smaller, or equivalently, it is the derivative of \(f(x)\) at \(x = 2\).

First, find the derivative of \(f(x)\):

\[
f'(x) = \frac{d}{dx} \left( -\frac{1}{2}x^2 + 2x \right) = -x + 2
\]

Now, evaluate the derivative at \(x = 2\):

\[
f'(2) = -2 + 2 = 0
\]

So, the slope of the tangent line at \(x = 2\) is \(0\).

---

### d) Use your estimate to find the equation of the tangent line at \(x = 2\):

The equation of the tangent line at \(x = 2\) can be written in point-slope form:

\[
y - y_1 = m(x - x_1)
\]

where \(m\) is the slope of the tangent line (which is \(0\)), and \((x_1, y_1)\) is the point of tangency, which is \((2, f(2)) = (2, 2)\).

Since the slope is \(0\), the equation of the tangent line is horizontal:

\[
y - 2 = 0(x - 2)
\]
\[
y = 2
\]

So, the equation of the tangent line at \(x = 2\) is \(y = 2\).

---

Would you like more details or have any questions?

Here are some related questions to further explore:
1. How does the derivative relate to the concept of a tangent line?
2. What is the geometric interpretation of the secant line?
3. Can you find the slope of the secant line for different intervals closer to \(x = 2\)?
4. How would the equation of the tangent line change if the slope was non-zero?
5. What are the implications of the tangent line having a slope of zero at \(x = 2\)?

**Tip:** The closer the two points on a secant line are, the more it approximates the tangent line at a point.

Find the slope of the secant lines over the specified intervals for the function f(x) = -(1/2)x^2 + 2x. a) [2, 3] b) [2, 2.1] Estimate the slope of the tangent line at x = 2. Use your estimate to find the equation of the tangent line at x = 2.

Learn how to find the slope of secant lines over the intervals [2, 3] and [2, 2.1], estimate the slope of the tangent line at x = 2, and derive the equation of the tangent line for the function f(x) = -(1/2)x^2 + 2x. Includes step-by-step calculations using calculus concepts like derivatives. Suitable for high school calculus students.

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