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How do you read the tangent slope of f(x)=2x3f(x)=2x^3 at any x-value using the complete derivative graph?

Once the full derivative curve is available, you select any desired xx-coordinate within the visible window, look vertically up or down to intersect the orange parabolic graph, and interpret that exact vertical height as the slope of the tangent line to the original blue curve at that same xx.

Conditions

  • The derivative curve spans the continuous plotting area.
  • Requires recognizing that the vertical position of the derivative graph encodes slope, not just the function output of ff itself.

Reasoning, step by step

  1. Choose an arbitrary target xx-value on the horizontal axis.
  2. Trace a vertical line from that xx-position up to the orange derivative curve.
  3. Read the corresponding yy-axis scale at the intersection point.
  4. Map that numerical height back to the steepness of the tangent line on the original cubic graph.

Example

The narrator picks x=−0.5x=-0.5 (which was not one of the seven initially adjusted markers), looks up to the orange curve, and visually estimates a slope slightly above 1, which analytically checks out to exactly 1.5 via 6(−0.5)26(-0.5)^2.

Common misconceptions

  • Mistaking derivative knowledge for only the seven specially manipulated sample points.
  • Confusing the height of the orange curve with the actual yy-value of the blue function f(x)f(x).

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