# 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

- ?Lv 72 months agoFavorite 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.

- billrussell42Lv 72 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 sLv 72 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

- llafferLv 72 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.