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#1 2005-12-28 05:48:29

katy
Member
Registered: 2005-12-28
Posts: 14

High School Student needs your help please...

hi everyone,I am a high school student and my exam is coming up around two weeks and I really need your help please....here are 2 math problems I really need answers...if anyone of you can solve any of them then I am really more than happy...
1)Find the equation of a parabola with vertex (-4,-1) and y-intercept -5
2)Find two numbers whose difference is 10 and whose product is a minimum

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#2 2005-12-28 06:46:30

God
Member
Registered: 2005-08-25
Posts: 59

Re: High School Student needs your help please...

1. General form of a parabola equation is: (y-k)=a*(x-h)^2 where (h,k) is the vertex and a is a scale factor.

That gives y+1 = a*(x+4)^2

Since (0,-5) works for the equation, substitute and solve for a:

-5 + 1 = a*(0+4)^2
-4 = a*16
a = -1/4

so your equation is:

y + 1 = -1/4 * (x+4)^2

2. Let x be one number and (x-10) be the other number...
    So you want to minimize x*(x-10), which is x^2 - 10*x. The lowest point is at the vertex, so the       
    x coordinate of the vertex is one number, and your other number is x-10.
    Vertex occurs at -b/2a = 10/2 = 5, so the two numbers are 5 and -5

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#3 2005-12-28 06:54:27

katy2
Guest

Re: High School Student needs your help please...

thanks for the answers, GOD...but i still do not understand this quote :"whose product is a minimum" of #2
what is a minimum?and how do you know whether the product is a min or max???

#4 2005-12-28 07:27:59

seerj
Member
Registered: 2005-12-21
Posts: 42

Re: High School Student needs your help please...

katy2 wrote:

thanks for the answers, GOD...but i still do not understand this quote :"whose product is a minimum" of #2
what is a minimum?and how do you know whether the product is a min or max???

U can decide is a  x1 is a minimum also by studying the graph of the function.
For example in a parabola with vertex in (-5,0)   -5 is a min! and max doesn't exist.

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#5 2005-12-28 07:57:27

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: High School Student needs your help please...

Well basicly its the same problem presented in two different forms. When you graph an equation or a function on a cartesian coordinate grid, the horizontal location on the x axis represents the value of what they call the independant variable. The height on the y axis represents the value of the expression evaluated at that location.

Therefore if you have the eqaution:

L - S = 10

And

S * L = Product

We can rearrange the first equation to find L = 10 + S and subsitute this expression for L:

S * (10 + S) = Product

10S + S^2 = Product

if we graph were to replace S with x and Product with Y we would have:

x^2 + 10x = y

this is the eqaution of a parabolla. This graph represents the product of the two numbers who's difference is 10. Therefore we can look at the low point on the graph to find the minimum product, and note what value of x is required. We used x to represent S so we can use the x coordinate of the lowpoint on the graph for S, and the solve for L. The equation y = x^2 + 10x is a parabolla that opens upward, thus the vertex is the lowpoint on the graph. Using any of the various methods to find the vertex, we find the x coordinate of the vertex is x = -5. We used  x to represent S so S = -5

Now lets use this value of S in the original equation to find L.

L - S = 10

L - (-5) = 10

L + 5 = 10

L = 5

So the two numbers are 5 and -5. :-)


A logarithm is just a misspelled algorithm.

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#6 2005-12-28 10:41:34

katy
Member
Registered: 2005-12-28
Posts: 14

Re: High School Student needs your help please...

mikau wrote:

Well basicly its the same problem presented in two different forms. When you graph an equation or a function on a cartesian coordinate grid, the horizontal location on the x axis represents the value of what they call the independant variable. The height on the y axis represents the value of the expression evaluated at that location.

Therefore if you have the eqaution:

L - S = 10

And

S * L = Product

We can rearrange the first equation to find L = 10 + S and subsitute this expression for L:

S * (10 + S) = Product

10S + S^2 = Product

if we graph were to replace S with x and Product with Y we would have:

x^2 + 10x = y

this is the eqaution of a parabolla. This graph represents the product of the two numbers who's difference is 10. Therefore we can look at the low point on the graph to find the minimum product, and note what value of x is required. We used x to represent S so we can use the x coordinate of the lowpoint on the graph for S, and the solve for L. The equation y = x^2 + 10x is a parabolla that opens upward, thus the vertex is the lowpoint on the graph. Using any of the various methods to find the vertex, we find the x coordinate of the vertex is x = -5. We used  x to represent S so S = -5

Now lets use this value of S in the original equation to find L.

L - S = 10

L - (-5) = 10

L + 5 = 10

L = 5

So the two numbers are 5 and -5. :-)

thank you so much for taking your time to help me solve the problems....this is extremely helpful to me....:-)

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#7 2005-12-28 10:50:24

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

Re: High School Student needs your help please...

Also, you could alternatively write
y= (x+5)(x-5) and solve for x, which is the average of the
two numbers you are looking for.
This might make the answer easier to see just by guessing.
Or continue.
y = x^2 - 25 and then logic tells you that x of zero makes y
turn out the smallest, because negative x's square to
positive numbers greater than the zero.


igloo myrtilles fourmis

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#8 2005-12-28 14:47:38

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: High School Student needs your help please...

thank you so much for taking your time to help me solve the problems....this is extremely helpful to me....:-)

No problem at all. I only wish high school chicks would ask for help with math in my neighborhood. ;-)

John E franklin, averaging the zero's, to find the axis of symetry. Why didn't I think of that? Just remember this is only ok to do when the parabolla is verticle. It cannot be solved for x instead and the xy coefficient must be zero. (remember the general conic equation)


A logarithm is just a misspelled algorithm.

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