Tangent line at a specific point — get the slope from the limit of the difference quotient.
A tangent line is a line, so its equation has the form $y = mx + b$ (or $y - y_1 = m(x - x_1)$). To write that equation you need two things: a slope and a point.
This problem hands you the point ($(2, 6)$) and asks you to find the line, so the only unknown is the slope. From Q1, you know the slope of a tangent line is the limit of the difference quotient as the two points collapse to one. So the recipe here is:
The only thing that's different from Q1 is the starting point: Q1 asked for the general slope (in terms of $x$), Q3 asks for the slope at a specific point. The algebra is the same, but $x$ gets pinned to $2$ in step 1.
Tangent slope at the specific point $x = 2$:
$$m_{\text{tan}} \;=\; \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$$This is just Q1's formula with $x$ replaced by $2$. Once you have the slope, write the tangent line in point-slope form:
$$y - 6 \;=\; m_{\text{tan}}\,(x - 2)$$Then expand and simplify to slope-intercept if the question asks for it.
Given $f(x) = x^2 + x$, point $(2,\ 6)$. First verify the point is on the curve: $f(2) = 4 + 2 = 6$ ✓.
Step 1 — Set up the difference quotient at $x = 2$.
$$m_{\text{tan}} \;=\; \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}$$Step 2 — Compute $f(2+h)$.
Substitute $(2 + h)$ for $x$ in $f(x) = x^2 + x$:
$$f(2+h) \;=\; (2+h)^2 + (2+h) \;=\; (4 + 4h + h^2) + (2 + h) \;=\; 6 + 5h + h^2$$Step 3 — Subtract $f(2) = 6$.
$$f(2+h) - f(2) \;=\; (6 + 5h + h^2) - 6 \;=\; 5h + h^2$$Step 4 — Divide by $h$ and factor.
$$\frac{5h + h^2}{h} \;=\; \frac{h(5 + h)}{h} \;=\; 5 + h$$Step 5 — Take the limit as $h \to 0$.
$$m_{\text{tan}} \;=\; \lim_{h \to 0} (5 + h) \;=\; 5 + 0 \;=\; 5$$Now we have the slope: $m = 5$. Use point-slope form with the point $(2,\ 6)$:
Step 6 — Point-slope form, then simplify.
$$y - 6 \;=\; 5(x - 2)$$ $$y - 6 \;=\; 5x - 10$$ $$y \;=\; 5x - 4$$Final answer:
Verification — does the line pass through $(2, 6)$?
$$y \;=\; 5(2) - 4 \;=\; 10 - 4 \;=\; 6 \quad ✓$$Good. The point we used in point-slope form is actually on the final line. (If it weren't, we'd have a sign error somewhere.)
Compare the difference quotient in Q1 vs Q3:
| Q1 (general slope) | $\lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}$ | answer is a function of $x$ |
| Q3 (slope at $x = 2$) | $\lim_{h \to 0} \dfrac{f(2+h) - f(2)}{h}$ | answer is a number |
| Q3 tangent line | $y - 6 = 5(x - 2)$ | slope + point form |
So the only mechanical change from Q1 to Q3 is: substitute the specific $x$-value into the difference quotient. The 5-step dance is otherwise identical. Once you have a slope, the tangent line is just the point-slope form from the "Slope and Equations of Lines" topic.
Write the equation of the tangent line to $f(x) = x^2 - 3x$ at the point $(1, -2)$.
Hint: same recipe. First confirm $(1, -2)$ is on the curve ($f(1) = ?$). Then build the difference quotient at $x = 1$, run the 5 steps, find the slope, use point-slope form, simplify.
Step 0 — Sanity: $f(1) = 1 - 3 = -2$ ✓ (point is on the curve)
Step 1: $m_{\text{tan}} = \lim_{h \to 0} \dfrac{f(1+h) - f(1)}{h}$
Step 2: $f(1+h) = (1+h)^2 - 3(1+h) = 1 + 2h + h^2 - 3 - 3h = -2 - h + h^2$
Step 3: $f(1+h) - f(1) = (-2 - h + h^2) - (-2) = -h + h^2$
Step 4: $\dfrac{-h + h^2}{h} = \dfrac{h(-1 + h)}{h} = -1 + h$
Step 5: $m_{\text{tan}} = \lim_{h \to 0}(-1 + h) = -1$
Step 6: $y - (-2) = -1(x - 1)$ → $y + 2 = -x + 1$ → $y = -x - 1$
Verify the line passes through $(1, -2)$: $y = -(1) - 1 = -2$ ✓. The slope is negative — from $(1, -2)$ going right, the line drops, which matches a downhill tangent on a parabola opening upward.
Send Q4 when ready. Q3 and Q4 will both use this 5-step recipe — the only thing that changes is the function $f$ and the point. Once you've done this dance a few times, the limit step will start to feel mechanical, and that's when the shortcut version (the power rule: $\frac{d}{dx}(x^2) = 2x$) will land as a "remember the dance, skip to the answer" trick. But not yet — keep doing the limit until it's automatic.