Precalc: help please?

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  • 4 months ago

    v(t) = t² + 11t - 4

    v(t) = ( t + 11/2 )² - ( 11/2 )²  - 4

    v(t) = ( t + 11/2 )² - 137/4

    vertex = ( -11/2, -137/4 )

    ━━━━━━━━━━━

    line of symmetry at vertex

    t = -11/2

    ━━━━

  • 4 months ago

    If you put this into vertex form:

    v(t) = a(t - h)² + k

    The vertex is the point (h, k)

    And the axis of symmetry would be t = h

    So to get your equation into vertex form:

    v(t) = t² + 11t - 4

    Let's get the right side into the form of (t² + bt) by adding 4 to both sides:

    v(t) + 4 = t² + 11t

    Now complete the square by adding 121/4 to both sides:

    v(t) + 4 + 121/4 = t² + 11t + 121/4

    Now we can simplify the left side and factor the right:

    v(t) + 16/4 + 121/4 = (t + 11/2)²

    v(t) + 137/4 = (t + 11/2)²

    Finally, solve for v(t) again:

    v(t) = (t + 11/2)² - 137/4

    The vertex is the point (-11/2, -137/4)

    And the axis of symmetry is t = -11/2

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