English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

Consider the quadratic relation y=2x²+12x=54.

a) Use two different methods to determine the equation in vertex form.

b) Which method(s) enable you to also find points on the graph that are equidistant from the axis of symmetry?

c) Which method reveals the x-intercepts of the graph.

d) If the graph of the relation is not required, is there an advantage to using one method over another? Explain your reasoning.

2006-07-24 11:04:23 · 6 answers · asked by many men 1 in Science & Mathematics Mathematics

6 answers

I'll list the different methods
1) Complete the square
2) Use the formula. The vertex occurs at x = -b/(2a)
3) Use calculus

2x^2 + 12x + 54
2(x^2 + 6x) + 54
2(x^2 + 6x + 9 - 9) + 54
2(x^2 + 6x + 9) - 2(9) + 54
y = 2(x+3)^2 + 36
vertex: (-3,36)

a = 2, b = 12
x = -b/2a
x = -12/(2*2)
x = -3
y = 2(-3)^2 + 12(-3) + 54
y = 36
vertex: (-3,36)

(b) and (c) Method 1
(d) A lot of people have trouble completing the square, so it may be easier to use method 2.
If you've learned calculus, then method 3 is the easiest and quickest.

2006-07-24 12:05:29 · answer #1 · answered by MsMath 7 · 1 1

Super POORLY POSED question, anyhow...

a)
If y=2x²+12x=54
then 2x²+12x-54=0

...and everything that "mathgirl826" and "Michael M" wrote is true except that the vertex is then at (-3,-72) and there are real roots at x=-9 and x=3.

To the solution methods already suggested I would add that completing the square using the general quadratic formula (Ref 1) also yields:

k=4ac-b²/4a

and that the roots are at

x1=h + sqrt(-k/a) and
x2=h - sqrt(-k/a)

Setting the first derivative of the original quadratic to zero yields

4x +12 = 0 or x=-3

...and the second derivative is positive- so it it concave up and symmetric around x= -3

b)
ANY method you use that gives you "h" gives you the axis of symmetry and thus the points equidistant from it.

c)
ANY method that gives an explicit relationship between "x" and "y" can be solved for its X-intercepts (roots). I suppose the "h" and "k" formulation is a little more elegant in that it highlights the underlying symmetry.

d)
I guess that depends on what IS required. :(

I think your instructor would rather be teaching philosophy...

I hope you will update this posting with what your instructor thought appropriate answers would be...

2006-07-24 23:31:47 · answer #2 · answered by Fred S 2 · 0 0

I'm going to assume that you meant y=2x^2+12x+54

a)
Vertex form -> y = a(x - h)^2 + k
First way, completing the square
y = 2x^2 + 12x + 54
y/2 = x^2 + 6x + 27
y/2 = x^2 + 6x + 9 +18
y/2 = (x+3)^2 + 18
y = 2(x+3)^2 + 36

Second way, quadratic equation (Not the best way)
y = 2x^2 + 12x + 54
0 = 2x^2 + 12x + 54 - y
x = (-12 +/- sqrt(144 - 4(2)(54-y))) / 4
4x = -12 +/- sqrt(144 - 4(2)(54-y))
4x + 12 = +/- sqrt(144 - 4(2)(54-y))
(4x + 12)^2 = 144 - 4(2)(54-y)
(4x + 12)^2 = 144 - (8)54 + 8y
(4x + 12)^2 = -288 + 8y
(16)(x + 3)^2 = -288 + 8y
2(x + 3)^2 = -36 + y
y = 2(x + 3)^2 + 36

b) I don't understand this question.
The vertical axis of symmetry is defined as y = 2(p)(x) + h where p is in the equation (x - h)^2 = 4p(y - k).
Therefore 4p = 1/2, p = 1/8
Axis of symmetry is y = 1/4x - 3

c) Solve for x when y = 0
y = 2x^2 + 12x + 54
0 = 2x^2 + 12x + 54
x = -12 +/- sqrt(144 - (4)(2)(54)) / 4
x = -12 +/- sqrt(-288) / 4
There are no real x-intercepts of this graph.

2006-07-24 18:41:05 · answer #3 · answered by Michael M 6 · 0 0

Is that supposed to be - 54 or +54?

2006-07-24 18:39:59 · answer #4 · answered by jenh42002 7 · 0 0

e) Go have a drink, thats a hard one.


Good Luck

2006-07-24 18:08:59 · answer #5 · answered by me 2 · 0 0

The equation is imaginary.....
Meaning that there is no solution....

2006-07-24 18:12:32 · answer #6 · answered by BK Randy 3 · 0 0

fedest.com, questions and answers