i need help with math work(reposted)?

**** Please list all answers clearly like: A) answer b)answer C)answer D)answer****

Without graphing, state whether each function has a maximum or a minimum. Then, write each

function in the form y = a(x - h)2 + k and find the minimum or maximum value and the value of x for which it occurs. (complete the square algebraically and using algebra tiles) /12T/I

a) y = 3x2 - 18x + 1

b) y = -4x2 - 32x - 11

c) y = -7x2 + 84x + 19

d) y = 4x2 - 20x + 7

 thank you so much!

4 Answers

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  • ?
    Lv 7
    2 months ago
    Favorite Answer

    // Note that all these functions are PARABOLAs.

    // If the leading term of the parabola is POSITIVE,

    //      the parabola is CONCAVE UP (U-shaped) so

    //      it won't have any maximum, but it will have a

    //      minimum at its vertex

    // If the leading term of the parabola is NEGATIVE,

    //      the parabola is CONCAVE DOWN (∩-shaped) so

    //      it won't have any minimum, but it will have a

    //      maximum at its vertex

    // If the parabola is given in standard form,

    //              f(x) = ax²+bx+c

    //     its vertex will be at (x,y) = (-b/2a, f(-b/2a))

    // If the parabola is written in vertex form,

    //              y = a(x-h)² + k

    //     where the vertex is (h,k)

    // To convert between the 2 equation forms,

    //     pay attention to the vertex:

    //              h = -b/2a     and    k = f(-b/2a)

    a) y = 3x² - 18x + 1

        LEADING TERM 3x² > 0 ⇒ CONCAVE UP

                                                ⇒ No maximum

                                                ⇒ Minimum at vertex (3, -26)

        VERTEX FORM y = 3(x-3)²-26

    ---------------------------

    b) y = -4x² - 32x - 11

        LEADING TERM -4x² < 0 ⇒ CONCAVE DOWN

                                                 ⇒ No minimum

                                                 ⇒ Maximum at vertex (-4, 53)

        VERTEX FORM y = -4(x+4)²+53

    ---------------------------

    c) y = -7x² + 84x + 19

        LEADING TERM -7x² < 0 ⇒ CONCAVE DOWN

                                                ⇒ No minimum

                                                ⇒ Maximum at vertex (6, 271)

        VERTEX FORM y = -7(x-6)²+271

    ---------------------------

    d) y = 4x² - 20x + 7

        LEADING TERM 4x² < 0 ⇒ CONCAVE UP

                                                ⇒ No maximum

                                                ⇒ Minimum at vertex (6, 271)

        VERTEX FORM y = -7(x-6)²+271

    --------------------------

    For verification, the graphs of the above parabolas have been provided.

    Attachment image
  • 2 months ago

    telling me exactly how to format my answers means you are too lazy to do that, and also means no answer from me. 

  • ted s
    Lv 7
    2 months ago

    why reposted ?? you certainly can complete the square...and min ; max ; max ; min...lead coefficient > 0 means min.....c) - 7 ( x - 6)² + 7 (6²) + 19 & x = 6

  • 2 months ago

    I'll show the first one and explain it.  The others are all pretty much the same.

    You want to get the values into vertex form:

    y = a(x - h)² + k

    This is a quadratic with the vertex at (h, k).  The value "k" is your min or max at the value x = h.  "k" is a minimum if "a" is positive and "k" is a maximum if "a" is negative.

    So the first one:

    y = 3x² - 18x + 1

    To prepare for completing the square, the right side has to be in the form (x² + bx)

    We can get the equation to this form by subtracting 1 from both sides, then dividing both sides by 3:

    y = 3x² - 18x + 1

    y - 1 = 3x² - 18x

    (y - 1) / 3 = x² - 6x

    Now we can complete the square by adding 9 to both sides:

    9 + (y - 1) / 3 = x² - 6x + 9

    Now we can factor the right side:

    9 + (y - 1) / 3 = (x - 3)²

    Finally, solve for y again keeping that binomial-squared in tact:

    (y - 1) / 3 = (x - 3)² - 9

    y - 1 = 3(x - 3)² - 27

    y = 3(x - 3)² - 26

    So here is your vertex form.  It tells us that y = -26 is the minimum at x = 3.

    We know it's a minimum since the "a" is positive.

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