Find the parabola with equation y = ax2 + bx whose tangent line at (1, 0) has equation y = 3x − 3.

Question

Find the parabola with equation y = ax² + bx , whose tangent line at (1, 0) has equation y = 3x − 3.

Answer 1

The problem could have stated find the equation of the tangent line at (1,0) for the function y = ax² + bx given that the tangent line is parallel to y = 3x + c.

From the fact that (1,0) is on the graph of the given function we have:

a + b = 0 (when x = 1, y=0)

y' = 2ax + b, but we know that the slope of the tangent line is 3 from y = 3x - 3 (or parallel to y = 3x + c).

y'(1) = 3 = 2a + b (when x = 1, y' = 3).

2a + b = 3

a + b = 0

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a = 3, b = -3

The target function is y = 3x² - 3x, (when x = 1, y = 0)

y' = 6x -3, when x = 1, y' = 3

Verify equation of tangent line: slope = 3, passing through (1,0)

y - 0 = 3(x - 1)

y = 3x - 3

Visit this graph and click on point P to see the parabola passing through (1,0) with a tangent line with equation y = 3x - 3 at that point.

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