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Find constants so a quadratic has a given tangent at a particular point

Let f be the quadratic function

    \[ f(x) = x^2 + ax + b \qquad a,b \text{ constants in } \mathbb{R}. \]

Find values for a and b such that the graph of f has a tangent line given by y = 2x at the point (2,4).


First, we compute the derivative

    \[ f'(x) = 2x + a. \]

So, if the line y = 2x is tangent to f at the point (2,4) we must have

    \[ f'(2) = 2 \quad \implies \quad 4+a = 2 \quad \implies \quad a = -2. \]

Next, the point (2,4) must be on the graph of f, i.e., we must have f(2) = 4. Thus,

    \[ f(2) = 4 \quad \implies \quad 4 + (-2)2 + b = 4 \quad \implies \quad b = 4. \]

So, the values are a = -2 and b = 4.

2 comments

  1. Anonymous says:

    I think there might be an error in this answer, why do you have f'(2) = 2?
    is asking for the the tangent to be equal to y = 2x. f'(x) = 2x + a, so at (2,4) a should = 0.
    and therefore b = 0 as well…

    • RoRi says:

      No, the answer is correct. It is asking for the tangent line to be equal to y = 2x at the point (2,4), not for the derivative to be 2x. (The derivative gives the slope of the tangent line, it is not a formula for the tangent line.) So, since y = 2x is tangent to the curve at (2,4) and this line has slope 2, we know the derivative must be equal to 2 at that point. Hence, f'(2) = 2. Does that make sense?

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