📝 How to find tangent line equation (18 MCQs)

📖 From Calculus • 3. The Derivation • 18 questions available

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What is How to find tangent line equation?

Definition:
To find the tangent line equation at a point, calculate the derivative to get the slope at that specific x-value, then use the point-slope form $y - y_1 = m(x - x_1)$ with the known point coordinates to construct the linear equation.

Example:
For $f(x) = x^2$ at $(1,1)$ , slope $m=2$ , so equation is $y - 1 = 2(x - 1)$ or $y = 2x - 1$ .

Reason:
This procedure combines differentiation with algebra, enabling students to approximate functions locally and solve optimization problems by identifying critical points where tangents are horizontal.

Easy

Medium

Hard

📝 All How to find tangent line equation MCQs

Q1. Given $f(x)=x^{3}-x$ and $x_{0}=1$ , what is the slope of the tangent line at this point?

A.0

B.2 ✅

C.-2

D.1

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The derivative is $f'(x)=3x^{2}-1$ . Substituting $x=1$ gives $f'(1)=3(1)^{2}-1=2$ . Hence the slope of the tangent line at $(1,f(1))$ is 2.

Q2. For $f(x)=x^{2}$ , compare the slopes of the tangent lines at $x=0$ and $x=2$ . Which statement is true?

A.The slopes are equal

B.Both slopes are zero

C.The slopes are different ✅

D.Both slopes are vertical

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The derivative $f'(x)=2x$ . At $x=0$ the slope is $0$ ; at $x=2$ the slope is $4$ . Since $0\neq4$ , the slopes are different.

Q3. If $f(x)=5x+3$ , what is the equation of the tangent line at any point on the graph?

y=5x+3

✅

y=5x+3+x_{0}

y=5x

y=3

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: A linear function’s graph coincides with its own tangent line at every point, so the tangent line equation is identical to the function: $y=5x+3$ .

Q4. For $f(x)=\sqrt{x}$ , at which $x$ -value does the tangent line have slope $\dfrac14$ ?

x=1

x=4

✅

x=16

x=9

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The derivative is $f'(x)=\dfrac{1}{2\sqrt{x}}$ . Setting $\dfrac{1}{2\sqrt{x}}=\dfrac14$ gives $\sqrt{x}=2$ and thus $x=4$ .

Q5. For $f(x)=x^{3}-3x$ , at which $x$ -values is the tangent line horizontal?

x=0

x=\pm1

✅

x=\pm2

D.No such points

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A horizontal tangent occurs when $f'(x)=0$ . Here $f'(x)=3x^{2}-3=3(x^{2}-1)$ . Solving $3(x^{2}-1)=0$ yields $x=\pm1$ .

Q6. If $f'(x)>0$ for every $x$ in an interval $(a,b)$ , which of the following must be true about the tangent lines on that interval?

A.All slopes are positive

B.The function is increasing

C.Both A and B ✅

D.None of the above

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: A positive derivative means each tangent line has a positive slope, which also indicates the function is strictly increasing on $(a,b)$ . Both statements are correct.

Q7. What is the point‑slope form of a line?

y-y_{1}=m(x-x_{1})

✅

y=mx+b

y=mx

x=mx+b

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The point‑slope form expresses a line through $(x_{1},y_{1})$ with slope $m$ as $y-y_{1}=m(x-x_{1})$ .

Q8. Let $g(x)=\ln x$ . Find the equation of the tangent line at $x_{0}=e$ .

y-1=\frac{1}{e}(x-e)

✅

y-1=\frac{e}{x}(x-e)

y=\frac{1}{e}x

y=1

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The derivative of $\ln x$ is $1/x$ ; at $x=e$ the slope is $1/e$ . The point is $(e,1)$ . Using point‑slope: $y-1=\frac{1}{e}(x-e)$ .

Q9. Compare the slope of the tangent line to $f(x)=x^{2}$ at $x=1$ with that of $h(x)=x^{3}$ at $x=1$ . Which is larger?

A.Slope of

f

is larger

B.Slope of

h

is larger ✅

C.The slopes are equal

D.Cannot be determined

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: $f'(x)=2x$ gives slope $2$ at $x=1$ . $h'(x)=3x^{2}$ gives slope $3$ at $x=1$ . Hence the tangent to $x^{3}$ is steeper.

Q10. For the function $h(x)=|x|$ , does the derivative exist at $x=0$ ?

A.Yes, and equals 0

B.No, it does not exist

C.Undefined ✅

D.It is infinite

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The absolute value function has a sharp corner at $x=0$ ; the left‑hand and right‑hand limits of the difference quotient differ, so the derivative is undefined there.

Q11. For $f(x)=x^{4}-4x^{2}$ , at which points does the tangent line pass through the origin?

x=0

x=\pm\frac{2}{\sqrt3}

✅

x=\pm1

D.No such points

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The condition $\frac{f(x)}{x}=f'(x)$ leads to $x^{3}-4x=4x^{3}-8x$ → $x(3x^{2}-4)=0$ . Excluding $x=0$ (division by zero) gives $x=\pm2/\sqrt3$ .

Q12. For $f(x)=e^{2x}$ , is there a point where the tangent line has a y‑intercept of 5?

x\approx0.1

x\approx-0.2

x\approx0.3

D.No such point ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The y‑intercept equals $e^{2x_{0}}(1-2x_{0})$ . This expression attains a maximum below 1, so it can never reach 5; thus no such point exists.

Q13. If a differentiable function $f$ has a tangent line at $x=a$ that passes through $(b,f(b))$ with $b\neq a$ , which theorem does this illustrate?

A.Mean Value Theorem

B.Rolle's Theorem

C.Intermediate Value Theorem ✅

D.Fundamental Theorem of Calculus

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The existence of a point $c$ between $a$ and $b$ where $f'(c)=\frac{f(b)-f(a)}{b-a}$ is precisely the statement of the Mean Value Theorem.

Q14. For $f(x)=\dfrac{1}{x}$ , find the tangent line at $x_{0}=2$ and state where this line meets the x‑axis.

x=0

x=4

✅

x=2

x=-4

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: $f'(x)=-1/x^{2}$ ; at $x=2$ the slope is $-1/4$ and the point is $(2,0.5)$ . The line $y-0.5=-\frac14(x-2)$ gives $y=0$ when $x=4$ .

Q15. What does $f'(x_{0})$ represent geometrically?

A.Slope of the tangent line ✅

B.Instantaneous rate of change

C.Area under the curve

D.Curvature of the graph

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The derivative at a point gives the slope of the line that just touches the curve at that point, i.e., the tangent line.

Q16. For $f(x)=x^{3}$ , what is the equation of the tangent line at $x_{0}=0$ ?

y=0

✅

y=x

y=3x

y=x^{3}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: $f'(x)=3x^{2}$ ; at $x=0$ the slope is 0 and the point is $(0,0)$ . Thus the tangent line is the horizontal line $y=0$ .

Q17. The tangent line to $y=\sin x$ at $x=\pi/6$ is given by $y=\frac12+\frac{\sqrt3}{2}\bigl(x-\frac{\pi}{6}\bigr)$ . Does this slope equal the derivative at that point?

A.true ✅

B.false

C.Cannot be determined

D.Insufficient information

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The derivative of $\sin x$ is $\cos x$ ; at $x=\pi/6$ , $\cos(\pi/6)=\sqrt3/2$ , which matches the slope in the given equation, so the statement is true.

Q18. A function has derivative $f'(x)=\dfrac{2x}{1+x^{2}}$ . Without finding $f$ , where are its tangent lines horizontal?

x=0

✅

B.No points

C.Where the denominator is zero

x=0

only

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: A horizontal tangent occurs when $f'(x)=0$ . The numerator $2x$ is zero only at $x=0$ ; the denominator never vanishes for real $x$ . Hence the only horizontal tangent is at $x=0$ .

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