2. Find the point on the curve y= x2 – 6x + 3 where the tangent is horizontal?

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  • 3 months ago

    ALGEBRAIC METHOD:

    You have a quadratic of the form:

    y = ax² + bx + c

    That function represents a parabola. The tangent will be horizontal at the vertex. And the x-coordinate of the vertex can be found using this formula:

    x = -b/(2a)

    (If you need help remembering that, it's essentially the quadratic formula but without the ±√ part)

    Plug in your values:

    x = -(-6) / (2*1)

    x = 6/2

    x = 3

    Finally calculate the y-value by plugging x=3 to the original function:

    y = 3² - 6(3) + 3

    y = 9 - 18 + 3

    y = -6

    Answer:

    (3, -6)

    CALCULUS METHOD:

    Original function:

    y = x² - 6x + 3

    The derivative represents the slope at each point, so take the derivative:

    y' = 2x - 6

    You want the tangent to be horizontal (slope = 0) so set that to 0.

    2x - 6 = 0

    2x = 6

    x = 6/2

    x = 3

    That gets you the x-coordinate. Plug in x=3 into the original equation to get the y-coordinate:

    y = 3² - 6(3) + 3

    y = 9 - 18 + 3

    y = -6

    Answer:

    (3, -6)

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