Business Calculus Quiz: Derivative As Slope
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Derivative As SlopeQuestion 1 of 12

The temperature T(t)T(t) in a chemical reaction follows T(t)=25+40e−0.2tT(t) = 25 + 40e^{-0.2t} degrees Celsius, where tt is time in minutes. What does the slope of the tangent line at t=5t = 5 represent in this context?

The temperature is decreasing at approximately 2.95°C per minute at the 5-minute mark
The temperature is increasing at approximately 2.95°C per minute at the 5-minute mark
The temperature will reach 2.95°C exactly 5 minutes after the reaction begins
The average rate of temperature change over the first 5 minutes is 2.95°C per minute
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Business Calculus Quiz

Business Calculus Quiz: Derivative As Slope

Practice Derivative As Slope in Business Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Derivative As Slope, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Calculus.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The temperature T(t)T(t) in a chemical reaction follows T(t)=25+40e−0.2tT(t) = 25 + 40e^{-0.2t} degrees Celsius, where tt is time in minutes. What does the slope of the tangent line at t=5t = 5 represent in this context?

  1. The temperature is decreasing at approximately 2.95°C per minute at the 5-minute mark (correct answer)
  2. The temperature is increasing at approximately 2.95°C per minute at the 5-minute mark
  3. The temperature will reach 2.95°C exactly 5 minutes after the reaction begins
  4. The average rate of temperature change over the first 5 minutes is 2.95°C per minute
Explanation: The slope of the tangent line at t = 5 is T'(5). First, T'(t) = 40(-0.2)e^(-0.2t) = -8e^(-0.2t). At t = 5: T'(5) = -8e^(-0.2×5) = -8e^(-1) = -8(0.368) ≈ -2.95. The negative sign indicates the temperature is decreasing, and the magnitude 2.95 represents the instantaneous rate of decrease in °C per minute. Choice B has the wrong sign, choice C misinterprets the derivative as a temperature value, and choice D confuses instantaneous rate with average rate.

Question 2

The profit P(x)P(x) in dollars from producing xx units of a product is given by the function P(x)=x3−18x2+96x+50P(x) = x^3 - 18x^2 + 96x + 50. The rate of change of profit is zero at two different production levels.

Based on the passage, what is the difference in profit between these two production levels?

  1. $4
  2. $32 (correct answer)
  3. $178
  4. $388
Explanation: The rate of change of profit is the derivative, P′(x)P'(x). We need to find where the slope of the tangent line is zero. P′(x)=3x2−36x+96P'(x) = 3x^2 - 36x + 96. Setting P′(x)=0P'(x) = 0 gives 3(x2−12x+32)=03(x^2 - 12x + 32) = 0, which factors as 3(x−4)(x−8)=03(x-4)(x-8) = 0. The two production levels are x=4x=4 and x=8x=8. Now we calculate the profit at each level: P(4)=43−18(42)+96(4)+50=64−288+384+50=210P(4) = 4^3 - 18(4^2) + 96(4) + 50 = 64 - 288 + 384 + 50 = 210. P(8)=83−18(82)+96(8)+50=512−1152+768+50=178P(8) = 8^3 - 18(8^2) + 96(8) + 50 = 512 - 1152 + 768 + 50 = 178. The difference in profit is ∣210−178∣=32|210 - 178| = 32.

Question 3

At what point (x,y)(x, y) on the curve f(x)=x3−6x2+5xf(x) = x^3 - 6x^2 + 5x is the tangent line parallel to the line y=−4x+10y = -4x + 10 and the yy-coordinate is negative?

  1. (1, 0)
  2. (3, -12) (correct answer)
  3. (2, -6)
  4. (-1, -12)
Explanation: The slope of the given line is −4-4. For the tangent line to be parallel, its slope must also be −4-4. The slope of the tangent line to f(x)f(x) is given by its derivative, f′(x)=3x2−12x+5f'(x) = 3x^2 - 12x + 5. We set the derivative equal to the desired slope: 3x2−12x+5=−43x^2 - 12x + 5 = -4. This simplifies to the quadratic equation 3x2−12x+9=03x^2 - 12x + 9 = 0, or x2−4x+3=0x^2 - 4x + 3 = 0. Factoring gives (x−1)(x−3)=0(x-1)(x-3) = 0, so the x-coordinates are x=1x=1 and x=3x=3. We find the corresponding y-coordinates: f(1)=13−6(1)2+5(1)=0f(1) = 1^3 - 6(1)^2 + 5(1) = 0. The point is (1,0)(1, 0). f(3)=33−6(3)2+5(3)=27−54+15=−12f(3) = 3^3 - 6(3)^2 + 5(3) = 27 - 54 + 15 = -12. The point is (3,−12)(3, -12). The question asks for the point where the y-coordinate is negative, which is (3,−12)(3, -12).

Question 4

The total cost to produce qq items is given by the function C(q)=0.1q2+5q+400C(q) = 0.1q^2 + 5q + 400. The average cost per item is A(q)=C(q)/qA(q) = C(q)/q.

For what quantity qq is the instantaneous rate of change of the cost equal to the average cost per item?

  1. 20
  2. 40
  3. 201020\sqrt{10} (correct answer)
  4. 401040\sqrt{10}
Explanation: The instantaneous rate of change of the cost is the derivative of the cost function, also known as the marginal cost, C′(q)C'(q). C′(q)=0.2q+5C'(q) = 0.2q + 5. The average cost per item is A(q)=C(q)q=0.1q2+5q+400q=0.1q+5+400qA(q) = \frac{C(q)}{q} = \frac{0.1q^2 + 5q + 400}{q} = 0.1q + 5 + \frac{400}{q}. We need to find the quantity qq where the instantaneous rate of change equals the average cost, so we set C′(q)=A(q)C'(q) = A(q): 0.2q+5=0.1q+5+400q0.2q + 5 = 0.1q + 5 + \frac{400}{q}. Subtracting 0.1q0.1q and 5 from both sides gives 0.1q=400q0.1q = \frac{400}{q}. Multiplying by qq gives 0.1q2=4000.1q^2 = 400. Dividing by 0.1 gives q2=4000q^2 = 4000. Taking the square root gives q=4000=400⋅10=2010q = \sqrt{4000} = \sqrt{400 \cdot 10} = 20\sqrt{10}.

Question 5

A function f(x)f(x) is differentiable at x=ax=a. The tangent line to the graph of f(x)f(x) at x=ax=a passes through the origin (0,0)(0,0). Assuming a≠0a \neq 0, which equation correctly describes the relationship between aa, f(a)f(a), and the derivative f′(a)f'(a)?

  1. f(a)=−a⋅f′(a)f(a) = -a \cdot f'(a)
  2. f′(a)=a⋅f(a)f'(a) = a \cdot f(a)
  3. f′(a)=f(a)f'(a) = f(a)
  4. f(a)=a⋅f′(a)f(a) = a \cdot f'(a) (correct answer)
Explanation: The equation of the tangent line to f(x)f(x) at x=ax=a can be written in point-slope form as y−f(a)=f′(a)(x−a)y - f(a) = f'(a)(x-a). The problem states that this line passes through the origin, (0,0)(0,0). We can substitute x=0x=0 and y=0y=0 into the equation: 0−f(a)=f′(a)(0−a)0 - f(a) = f'(a)(0-a). This simplifies to −f(a)=f′(a)(−a)-f(a) = f'(a)(-a), which is −f(a)=−a⋅f′(a)-f(a) = -a \cdot f'(a). Multiplying both sides by -1 gives the relationship f(a)=a⋅f′(a)f(a) = a \cdot f'(a).

Question 6

The line y=5x−7y = 5x - 7 is tangent to the graph of a differentiable function f(x)f(x) at the point (2,3)(2, 3). What is the value of lim⁡h→0f(2+h)−f(2)h\lim_{h \to 0} \frac{f(2+h)-f(2)}{h}?

  1. -7
  2. 2
  3. 3
  4. 5 (correct answer)
Explanation: The expression lim⁡h→0f(2+h)−f(2)h\lim_{h \to 0} \frac{f(2+h)-f(2)}{h} is the definition of the derivative of f(x)f(x) at x=2x=2, which is f′(2)f'(2). The derivative of a function at a point gives the slope of the tangent line at that point. The problem states that the line y=5x−7y = 5x - 7 is tangent to the graph of f(x)f(x) at x=2x=2. The slope of this line is 5. Therefore, f′(2)=5f'(2) = 5.

Question 7

What is the x-intercept of the line tangent to the graph of the function f(x)=xln⁡(x)f(x) = x \ln(x) at the point where x=e2x=e^2?

  1. e2/3e^2/3 (correct answer)
  2. e2e^2
  3. −e2-e^2
  4. −e2/3-e^2/3
Explanation: First, find the point of tangency. At x1=e2x_1 = e^2, the y-coordinate is y1=f(e2)=e2ln⁡(e2)=e2(2)=2e2y_1 = f(e^2) = e^2 \ln(e^2) = e^2(2) = 2e^2. The point is (e2,2e2)(e^2, 2e^2). Next, find the slope of the tangent line by taking the derivative using the product rule: f′(x)=(1)ln⁡(x)+x(1x)=ln⁡(x)+1f'(x) = (1)\ln(x) + x(\frac{1}{x}) = \ln(x) + 1. The slope at x=e2x=e^2 is m=f′(e2)=ln⁡(e2)+1=2+1=3m = f'(e^2) = \ln(e^2) + 1 = 2 + 1 = 3. The equation of the tangent line is y−2e2=3(x−e2)y - 2e^2 = 3(x - e^2). To find the x-intercept, set y=0y=0: 0−2e2=3(x−e2)0 - 2e^2 = 3(x - e^2). This simplifies to −2e2=3x−3e2-2e^2 = 3x - 3e^2. Adding 3e23e^2 to both sides gives e2=3xe^2 = 3x, so x=e2/3x = e^2/3.

Question 8

The line y=8x+by = 8x + b is tangent to the curve y=2x2y = 2x^2. What is the value of the y-intercept, bb?

  1. -8 (correct answer)
  2. 0
  3. 8
  4. 16
Explanation: The slope of the tangent line to the curve y=2x2y=2x^2 is given by its derivative, y′=4xy' = 4x. The given tangent line is y=8x+by=8x+b, which has a slope of 8. To find the point of tangency, we set the derivative equal to the slope: 4x=84x = 8, which gives x=2x=2. Now we find the y-coordinate of this point on the curve: y=2(22)=8y = 2(2^2) = 8. So, the point of tangency is (2,8)(2, 8). Since this point must lie on the tangent line y=8x+by=8x+b, we substitute its coordinates into the line's equation: 8=8(2)+b8 = 8(2) + b. This gives 8=16+b8 = 16 + b, so b=−8b = -8.

Question 9

The graphs of f(x)=x2+ax+bf(x) = x^2 + ax + b and g(x)=−x2+1g(x) = -x^2 + 1 are tangent to each other at the point (1,0)(1, 0). What is the value of aa?

  1. -4 (correct answer)
  2. -2
  3. 2
  4. 3
Explanation: For the two curves to be tangent at a point, they must both pass through that point and have the same slope (the same derivative) at that point. The point (1,0)(1, 0) is on both curves. For f(x)f(x), this means f(1)=12+a(1)+b=0f(1) = 1^2 + a(1) + b = 0, so 1+a+b=01+a+b=0. More importantly, their derivatives must be equal at x=1x=1. The derivatives are f′(x)=2x+af'(x) = 2x + a and g′(x)=−2xg'(x) = -2x. Setting them equal at x=1x=1: f′(1)=g′(1)  ⟹  2(1)+a=−2(1)f'(1) = g'(1) \implies 2(1) + a = -2(1). This simplifies to 2+a=−22+a = -2, which gives a=−4a=-4.

Question 10

A line that is tangent to the parabola f(x)=x2+1f(x) = x^2 + 1 at the point where x=ax=a also passes through the external point (0,−3)(0, -3). What is the value of a2a^2?

  1. 1
  2. 2
  3. 4 (correct answer)
  4. 5
Explanation: The derivative of the function is f′(x)=2xf'(x) = 2x. At x=ax=a, the slope of the tangent line is m=f′(a)=2am = f'(a) = 2a. The point of tangency on the parabola is (a,f(a))(a, f(a)), which is (a,a2+1)(a, a^2+1). The equation of the tangent line is y−(a2+1)=2a(x−a)y - (a^2+1) = 2a(x-a). Since this line passes through the point (0,−3)(0, -3), we can substitute these values for xx and yy: −3−(a2+1)=2a(0−a)-3 - (a^2+1) = 2a(0-a). Simplifying the equation gives −4−a2=−2a2-4 - a^2 = -2a^2. Adding 2a22a^2 to both sides gives a2−4=0a^2 - 4 = 0, so a2=4a^2 = 4.

Question 11

For the function f(x)=13x3−32x2+5f(x) = \frac{1}{3}x^3 - \frac{3}{2}x^2 + 5, the tangent line to its graph is perpendicular to the line x+4y=12x + 4y = 12 at two distinct x-values. What is the sum of these x-values?

  1. -4
  2. -3
  3. 3 (correct answer)
  4. 4
Explanation: First, find the slope of the given line x+4y=12x + 4y = 12. Rearranging to slope-intercept form gives 4y=−x+124y = -x + 12, so y=−14x+3y = -\frac{1}{4}x + 3. The slope is m1=−1/4m_1 = -1/4. The slope of a perpendicular line is the negative reciprocal, m2=−1/(−1/4)=4m_2 = -1/(-1/4) = 4. The slope of the tangent line to f(x)f(x) is given by its derivative, f′(x)=x2−3xf'(x) = x^2 - 3x. We set the derivative equal to the required perpendicular slope: x2−3x=4x^2 - 3x = 4. This gives the quadratic equation x2−3x−4=0x^2 - 3x - 4 = 0, which factors as (x−4)(x+1)=0(x-4)(x+1) = 0. The two x-values are x=4x=4 and x=−1x=-1. Their sum is 4+(−1)=34 + (-1) = 3.

Question 12

Consider the function h(x)=x2+16h(x) = \sqrt{x^2 + 16}. What is the geometric meaning of h′(4)h'(4)?

  1. It represents the slope of the secant line from the origin to the point (4,h(4))(4, h(4))
  2. It represents the slope of the tangent line to the curve at the point (4,42)(4, 4\sqrt{2}) (correct answer)
  3. It represents the average rate of change of h(x)h(x) over the interval [0,4][0, 4]
  4. It represents the slope of the line connecting (4,h(4))(4, h(4)) to (0,h(0))(0, h(0))
Explanation: h'(4) represents the instantaneous rate of change of h(x) at x = 4, which is geometrically interpreted as the slope of the tangent line to the curve at that point. First, let's find h(4) = √(16 + 16) = √32 = 4√2. The derivative is h'(x) = x/√(x² + 16), so h'(4) = 4/√32 = 4/(4√2) = 1/√2 = √2/2. This is the slope of the tangent line at (4, 4√2). Choice A describes a secant line, choice C describes average rate of change, and choice D describes another secant line.