Anonymous
Anonymous asked in Education & ReferenceHomework Help · 10 years ago

Find k such that the line y = 5x + 3 is tangent to the graph of the function below. y=ksqrt(x)?

what does k=?

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  • 10 years ago
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    y = k√x

    y' = k / 2√x

    Now if the given line is tangent to y = k√x

    then the slope of the line equals k / 2√x

    so k / 2√x = 5

    k = 10√x [eqn 1]

    Now to find the point where tangent touches the curve:

    k√x = 5x + 3 [eqn 2]

    Subs [eqn 1] into [eqn 2] → 10√x . √x = 5x + 3

    10x = 5x + 3

    5x = 3

    x = 3/5

    Now subs x = 3/5 into [eqn 1] → k = 10√(3/5)

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