Points and gradients identify coefficients

A curve CC has equation y=a(2x+1)2+bx+ce2xy=a(2x+1)^2+bx+ce^{2x}, where a,b,ca,b,c are constants. It is given that CC passes through the point (1,17−e2)(1,17-e^2) and the gradient of CC at the point (0,1)(0,1) is 5. Find the equation of CC. marks

[2021/TMJC/Prelim I/1]

Solution

Q

Finding coefficients

A curve CC has equation y=a(2x+1)2+bx+ce2xy=a(2x+1)^2+bx+ce^{2x}, where a,b,ca,b,c are constants. It is given that CC passes through the point (1, 17−e2)(1,\,17-e^2) and that the gradient of CC at the point (0,1)(0,1) is 5. Find the equation of CC.

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Hint

  1. Count the conditions: a point on CC (twice, since (0,1)(0,1) is also on CC) and a gradient. That is three equations.

  2. The gradient needs dydx\frac{dy}{dx}, so differentiate first.

  3. Solve the linear system in a,b,ca,b,c.

Answer · Q

y=2(2x+1)2−x−e2xy=2(2x+1)^2-x-e^{2x}.

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Three conditions on (a, b, c)

The unknowns a, b, c are the axes. Each condition is a plane.

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The fitted curve, its point and its tangent

C passes through (0, 1) with gradient 5 (tangent y = 5x + 1) and through (1, 17 − e²).

Switch on a feature to add it to the sketch and see how it is found. A full-mark sketch shows all of them.

  • Fitted curve C
  • Its tangent at (0, 1)
Ideas to try
Is C above or below its tangent near x=0x=0?