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The Slope of a Tangent Line to a Curve. or The slope at a Point.

The Slope of a Tangent Line to a Curve. or The slope at a Point. Answer question #1 The tangent line is the line that includes the point on the graph and has the same slope as the point on the graph. f (x) or y. x. lets find the slope of the curve at that point.

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The Slope of a Tangent Line to a Curve. or The slope at a Point.

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  1. The Slope of a Tangent Line to a Curve.orThe slope at a Point.

  2. Answer question #1 The tangent line is the line that includes the point on the graph and has the same slope as the point on the graph.

  3. f(x) or y x lets find the slope of the curve at that point I want to find the slope of the tangent or purple line, why? f (x) anchor (2, 1) 2) f(x)=0.25x2 (x, f(x)) f(2) = 1 x = 2 start by finding the slope of the secant line by choosing a 2nd point

  4. (x+h, f(x+h))or(x+x, f(x+x)) f(x+h) or f(x+x) f(4) = 4 h or x x+h or x+x select a second point and draw the secant line (4, 4) f (x) f(x)=0.25x2 (x, f(x)) (2, 1) h = x =2 2+h = 4

  5. x=2 f (x) (4, 4) =0.25x2 (x+x, f(x+x)) slope of secant = (2, 1) (x, f(x))

  6. x x=1.5 f (x) (?, ?) =0.25x2 (x+x, f(x+x)) slope of secant = (2, 1) (x, f(x))

  7. x x = 1 f (x) (?, ?) =0.25x2 (x+x, f(x+x)) slope of secant = (2, 1) (x, f(x))

  8. x=0.25 x f (x) (?, ?) =0.25x2 (x+x, f(x+x)) slope of secant = (2, 1) (x, f(x))

  9. x f (x) =0.25x2 (x+x, f(x+x)) slope of secant = (x, f(x)) What is happening to x? What is this process called?

  10. (x+x, f(x+x)) f(x+x) h or x so at (2,1) the tangent slope is 0.5(2) = 1 x+x f(x)=0.25x2 (4, 4) (2, 1) (x, f(x)) f(4) = 4 2+h = 4

  11. at (-1, 2) m = 2(-1) or -2 at (0, 1) m = 2(0) or 0

  12. Use the definition of the derivative to find f ‘(x) The derivative only exists if the function is continuous at the point.

  13. Find an equation of the tangent line to the graph of f at the point then use a calculator to graph the function and its tangent line at the point to check your answer.

  14. Find the equations of the 2 tangent lines to the graph of f that pass through the point. 1- Find the slope of the tangent (the derivative) 2- Find the slope of the point and (x, y) and set it equal to the derivative. Solve. (remember y = x2) 3- Plug your solutions into the derivative to find the slopes of the tangent lines – you now have a slope/point, find the equations

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