1. Given f(x) = 2x 2 − x + 2, use the limit definition m = lim h→0 f(a + h) − f(a) h or limx→a f(x) − f(a) x − a to find the slope of the tangent line to the curve at x = 2. Use your result to write an equation of the tangent line at x = 2

We are asked to find the slope of the tangent line to the curve \( f(x) = 2x^2 - x + 2 \) at \( x = 2 \), and then use that result to write the equation of the tangent line. To do this, we'll use the limit definition of the derivative:

\[
m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}
\]

where \( a = 2 \). Let's break this process into steps.

### Step 1: Compute \( f(a + h) \)
First, substitute \( x = a + h \) into \( f(x) \). Here, \( a = 2 \), so we find \( f(2 + h) \):

\[
f(2 + h) = 2(2 + h)^2 - (2 + h) + 2
\]

Simplifying the right-hand side:

\[
= 2(4 + 4h + h^2) - (2 + h) + 2
\]
\[
= 2(4 + 4h + h^2) - 2 - h + 2
\]
\[
= 8 + 8h + 2h^2 - 2 - h + 2
\]
\[
= 8h + 2h^2 + 8 - 2 - h + 2
\]
\[
= 2h^2 + 7h + 8
\]

So, \( f(2 + h) = 2h^2 + 7h + 8 \).

### Step 2: Compute \( f(a) \)
Next, compute \( f(2) \):

\[
f(2) = 2(2)^2 - 2 + 2 = 8 - 2 + 2 = 8
\]

### Step 3: Apply the limit definition of the derivative
Now, substitute \( f(2 + h) \) and \( f(2) \) into the limit definition:

\[
m = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h}
\]
\[
= \lim_{h \to 0} \frac{(2h^2 + 7h + 8) - 8}{h}
\]
\[
= \lim_{h \to 0} \frac{2h^2 + 7h}{h}
\]
\[
= \lim_{h \to 0} (2h + 7)
\]

As \( h \to 0 \), the expression becomes:

\[
m = 7
\]

So, the slope of the tangent line at \( x = 2 \) is \( 7 \).

### Step 4: Write the equation of the tangent line
To find the equation of the tangent line, we use the point-slope form of a line:

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

Here, \( m = 7 \), \( x_1 = 2 \), and \( y_1 = f(2) = 8 \). Substituting these values:

\[
y - 8 = 7(x - 2)
\]

Simplifying:

\[
y - 8 = 7x - 14
\]
\[
y = 7x - 6
\]

Thus, the equation of the tangent line at \( x = 2 \) is:

\[
y = 7x - 6
\]

Would you like a detailed breakdown of any part, or any clarifications?

Here are 5 related questions for further practice:
1. How would you find the slope of the tangent line for \( f(x) = 3x^2 - 4x + 1 \) at \( x = 1 \)?
2. What does the derivative of a function tell us geometrically?
3. Can you find the derivative of \( f(x) = x^3 - 2x + 5 \) using the limit definition?
4. How would the tangent line at \( x = 0 \) differ from the one at \( x = 2 \) for the same function?
5. How do the limits in calculus relate to the concept of continuity?

**Tip**: When finding the equation of a tangent line, always double-check both the slope and the point of tangency, as these are critical in the point-slope form of the line.

Given f(x) = 2x^2 − x + 2, use the limit definition m = lim h→0 f(a + h) − f(a) / h or lim x→a f(x) − f(a) / (x − a) to find the slope of the tangent line to the curve at x = 2. Use your result to write an equation of the tangent line at x = 2.

Discover how to use the limit definition of the derivative to find the slope of the tangent line to the curve f(x) = 2x² − x + 2 at x = 2. This detailed solution also covers how to write the equation of the tangent line using the point-slope formula. Suitable for high school calculus students.

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