← Back to 2. Limits and Continuity an Introduction

📝 Tangent line problem and slope (14 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 14 questions available

← Previous Topic📝 Tangent line problem and slopeArea problem and limits:Next Topic→

What is Tangent line problem and slope?

Definition:
The tangent line problem involves finding the slope of a line that touches a curve at a single point, representing the instantaneous rate of change. This slope is defined as the limit of the average rate of change over intervals as the interval shrinks to zero, formally $m_{\text{tan}} = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$ . This foundational concept bridges algebra and calculus by generalizing the slope of a secant line.

Example:
Find the slope of the tangent line to $f(x) = x^2$ at $x = 3$ .
Solution: $m = \lim_{h \to 0} \frac{(3+h)^2 - 9}{h} = \lim_{h \to 0} \frac{9+6h+h^2-9}{h} = \lim_{h \to 0} (6+h) = 6$ .

Reason:
This problem is essential because it defines the derivative, which is the core tool for analyzing rates of change in physics, economics, and engineering, enabling predictions of motion, growth, and optimization.

Easy

Medium

Hard

📝 All Tangent line problem and slope MCQs

Q1. Given $f(x)=x^{2}$ and the point $P(3,9)$ on its graph, which statement correctly relates the slope of the tangent line at $P$ to the derivative $f'(3)$ ?

A.The slope equals

f&#039;(3)

. ✅

B.The slope is twice

f&#039;(3)

C.The slope is half of

f&#039;(3)

D.The slope is unrelated to

f&#039;(3)

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Since the slope of the tangent line at a point is precisely the derivative of the function at that point, the correct relationship is that the slope equals $f'(3)$ . For $f(x)=x^{2}$ , $f'(x)=2x$ , giving a slope of $6$ at $x=3$ .

Q2. If a function $g$ satisfies $g'(x)=3x^{2}-4$ and its tangent line at $x=2$ passes through $(2,g(2))$ , which expression gives the y‑intercept of that tangent line?

g(2)-8

g(2)-12

g(2)-16

g(2)-20

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The slope at $x=2$ is $g'(2)=3(2)^{2}-4=8$ . Using point‑slope form, the line is $y-g(2)=8(x-2)$ . Setting $x=0$ yields the y‑intercept $y=g(2)-8\cdot2=g(2)-16$ . Hence the correct expression is $g(2)-16$ .

Q3. For a differentiable function $h$ , the tangent line at $x=a$ is horizontal. Which inference must be true?

h

has a local extremum at

a

. ✅

h

is constant on an interval containing

a

h&#039;(a)=0

h&#039;&#039;(a)=0

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A horizontal tangent means the line has zero slope, which is exactly the definition of the derivative being zero at that point: $h'(a)=0$ . While a zero derivative can indicate a local extremum, it does not guarantee one, nor does it imply constancy or a zero second derivative. Thus only $h'(a)=0$ is certain.

Q4. Consider $f(x)=\sin x$ . Which $x_{0}$ in $(0,\pi)$ makes the tangent line at $(x_{0},\sin x_{0})$ pass through $(\pi,0)$ ?

x_{0}\approx 0.86

x_{0}\approx 1.57

✅

x_{0}\approx 2.00

x_{0}\approx 2.68

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The tangent line at $x_{0}$ is $y=\sin x_{0}+\cos x_{0}(x-x_{0})$ . Substituting $(\pi,0)$ gives $0=\sin x_{0}+\cos x_{0}(\pi-x_{0})$ . Solving numerically on $(0,\pi)$ yields $x_{0}\approx1.57$ (essentially $\pi/2$ ), where $\sin x_{0}=1$ and $\cos x_{0}=0$ , satisfying the equation. Hence the correct choice is $x_{0}\approx1.57$ .

Q5. Compare the slopes of the tangent lines to $f(x)=x^{3}$ at $x=1$ and $x=-1$ . Which is true?

A.The slope at

x=1

is larger.

B.The slope at

x=-1

is larger.

C.The slopes are equal. ✅

D.Both slopes are zero.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The derivative of $f(x)=x^{3}$ is $f'(x)=3x^{2}$ . Evaluating at $x=1$ and $x=-1$ gives $f'(1)=3$ and $f'(-1)=3$ . Since both slopes are the same, the correct statement is that the slopes are equal, which corresponds to option C. (Option B is intentionally incorrect to test comparison.)

Q6. Which function has the tangent line at $x=0$ with the greatest y‑intercept?

f_{1}(x)=x+2

✅

f_{2}(x)=x^{2}

f_{3}(x)=\sin x

f_{4}(x)=e^{x}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: At $x=0$ the tangent line equals the function’s value plus its derivative times $x$ . For $f_{1}$ , the line is $y=2+x$ (intercept 2). For $f_{2}$ , the line is $y=0$ . For $f_{3}$ , the line is $y=x$ (intercept 0). For $f_{4}$ , the line is $y=1+x$ (intercept 1). Thus $f_{1}$ yields the largest intercept.

Q7. Which statement correctly describes the relationship between the tangent line to $y=|x|$ at $x=0$ and the secant line joining $(-1,1)$ and $(1,1)$ ?

A.The tangent line has slope 0, same as the secant line.

B.The tangent line does not exist at

x=0

. ✅

C.The tangent line coincides with the secant line.

D.Both lines are vertical.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The absolute‑value function $y=|x|$ has a sharp corner at the origin, so its derivative does not exist there. Consequently, a tangent line at $x=0$ cannot be defined, whereas the secant line through $(-1,1)$ and $(1,1)$ is horizontal. Hence the correct description is that the tangent line does not exist.

Q8. When a constant $k$ is added to a function $p(x)$ , how does the tangent line at a given point change?

A.The slope remains the same, and the y‑intercept increases by

k

. ✅

B.Both slope and intercept increase by

k

C.The slope increases by

k

, intercept unchanged.

D.Neither slope nor intercept changes.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Adding a constant shifts the entire graph vertically without affecting its shape. The derivative, which gives the slope of the tangent line, is unchanged, so the slope stays the same. The vertical shift raises every point, including the intercept, by exactly $k$ . Hence option A is correct.

Q9. To find the equation of the tangent line to $y=x^{3}$ at the point $(2,8)$ , which formula should be applied?

y-8 = 3(2)^{2}(x-2)

✅

y-8 = (x-2)^{3}

y = 8 + 3(x-2)

y = 8x^{2}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The tangent line uses the point‑slope form $y-y_{0}=f'(x_{0})(x-x_{0})$ . For $f(x)=x^{3}$ , $f'(x)=3x^{2}$ . At $x_{0}=2$ , the slope is $3(2)^{2}=12$ . Substituting gives $y-8=12(x-2)$ , which matches option A. The other options misuse the derivative or algebraic form.

Q10. If two distinct points on a curve share the same tangent slope, what does this tell us about the derivative of the function?

A.The derivative takes the same value at those points. ✅

B.The function is linear between the points.

C.The function has a constant second derivative.

D.The function must be periodic.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The derivative of a function at a point equals the slope of the tangent line there. When two points have identical tangent slopes, the derivative evaluates to the same number at both locations. This does not imply linearity or any property beyond equality of the first derivative values. Hence option A is correct.

Q11. When a function satisfies $f'(x)=0$ for every $x$ in an interval, which statement about its tangent lines on that interval is accurate?

A.All tangent lines are vertical.

B.All tangent lines are horizontal. ✅

C.All tangent lines coincide with the x‑axis.

D.The function has no tangent lines.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A zero derivative indicates that the slope of the tangent line is zero throughout the interval, meaning each tangent line is horizontal. A horizontal line is parallel to the x‑axis, but not necessarily coincident with it unless the function’s value is zero. Thus option B correctly captures the situation.

Q12. If the tangent line to a curve at point $P$ meets the curve again at point $Q$ , under what condition does $Q$ coincide with $P$ ?

A.When the curvature at

P

is zero.

B.When the derivative at

P

is zero.

C.When the curve is a straight line. ✅

D.When the function is quadratic.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: If the curve itself is a straight line, every tangent line coincides with the curve, so the only intersection point is the point of tangency $P$ itself; thus $Q=P$ . For any non‑linear curve, a tangent line typically meets the curve at another distinct point (or not at all). Hence the condition is that the curve be linear, option C.

Q13. What is the formal definition of the tangent line to a curve at a given point?

A.The limit of secant lines as the second point approaches the first. ✅

B.A line that touches the curve at exactly one point.

C.A line perpendicular to the radius vector at that point.

D.The line with the same slope as the curve at that point.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The precise definition states that the tangent line is the limit of the secant lines through the point of tangency and a second point on the curve as the second point approaches the first. This captures the notion of “instantaneous” direction and distinguishes it from merely touching the curve. Option A reflects this definition.

Q14. The derivative $f'(x_{0})$ represents which geometric quantity for the graph of $y=f(x)$ at $x_{0}$ ?

A.The slope of the tangent line at

(x_{0},f(x_{0}))

. ✅

B.The curvature of the graph at that point.

C.The distance from the origin to the point.

D.The area under the curve up to

x_{0}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: By definition, the derivative at a point gives the instantaneous rate of change of the function, which geometrically is the slope of the line that just touches the curve at that point—the tangent line. Therefore the correct geometric interpretation is option A.

🔗 Related Topics (MCQs)

📝 Area problem and limits MCQs 📝 Limits intuitive understanding MCQs 📝 Left hand and right hand limits MCQs 📝 One sided vs two sided limits MCQs 📝 Infinite limits and unbounded behavior MCQs

← Previous Topic📝 Tangent line problem and slopeArea problem and limits:Next Topic→