# Parabola question.... exam ..plz help asap..

• Jan 20th 2008, 05:29 PM
ruscutie100
Parabola question.... exam ..plz help asap..
Write an equation of a quadratic functon that satisfies each set of condtions.

a) the function has a minimum value of 6 at x=4.

b) The parabola is congruent to y = 3x^2 and has a minimum value of 7.

the answer for part a) y = 3(x-4)^2 + 6
b) y = 3x^2+ 7

these answers are from the back of my book.. it also says that "Answers may vary"... plz help me.. .thnx(Doh)(Doh):(:(:confused:
• Jan 21st 2008, 12:13 AM
Jhevon
Quote:

Originally Posted by ruscutie100
Write an equation of a quadratic functon that satisfies each set of condtions.

a) the function has a minimum value of 6 at x=4.

b) The parabola is congruent to y = 3x^2 and has a minimum value of 7.

the answer for part a) y = 3(x-4)^2 + 6
b) y = 3x^2+ 7

these answers are from the back of my book.. it also says that "Answers may vary"... plz help me.. .thnx(Doh)(Doh):(:(:confused:

• Jan 21st 2008, 02:46 AM
mr fantastic
Quote:

Originally Posted by ruscutie100
Write an equation of a quadratic functon that satisfies each set of condtions.

a) the function has a minimum value of 6 at x=4.

b) The parabola is congruent to y = 3x^2 and has a minimum value of 7.

the answer for part a) y = 3(x-4)^2 + 6
b) y = 3x^2+ 7

these answers are from the back of my book.. it also says that "Answers may vary"... plz help me.. .thnx(Doh)(Doh):(:(:confused:

a) The (minimum) turning point is at (4, 6). You're familiar with the turning point form of a parabola - $y = a(x - h)^2 + k$ where the turning point is at (h, k) - right?

So any answer of the form $y = a(x-4)^2 + 6, \, a > 0, \,$ satisfies the given property. The book chose a = 3. Other concrete answers can be got by choosing a = 1, 2, 4, 1/2, ......

b) The turning point has y-coordinate y = 7. So k = 7 in the turning point form. In the turning point form, a 'controls the shape', so a = 3.

So any answer of the form $y = 3(x - h)^2 + 7 \,$ satisfies the given properties. The book chose h = 0, probably the simplest answer. Other concrete answers can be got by choosing h = 1, 2, 3, -1, -2 -3, 1/2, -1/2, ......
• Jan 21st 2008, 02:47 AM
mr fantastic
Quote:

Originally Posted by Jhevon

(Rofl) That's because the answers are taken from the back of the book! (Doh)
• Jan 21st 2008, 07:01 PM
Jhevon
Quote:

Originally Posted by mr fantastic
(Rofl) That's because the answers are taken from the back of the book!

ah, yes, i see :D

Quote:

(Doh)
that's my line!
• Jan 21st 2008, 09:11 PM
mr fantastic
Quote:

Originally Posted by Jhevon
[snip]

that's my line!

Ah yes. But I learnt it from you :D
• Jan 25th 2008, 02:38 PM
ruscutie100
Quote:

Originally Posted by mr fantastic
a) The (minimum) turning point is at (4, 6). You're familiar with the turning point form of a parabola - $y = a(x - h)^2 + k$ where the turning point is at (h, k) - right?

So any answer of the form $y = a(x-4)^2 + 6, \, a > 0, \,$ satisfies the given property. The book chose a = 3. Other concrete answers can be got by choosing a = 1, 2, 4, 1/2, ......

b) The turning point has y-coordinate y = 7. So k = 7 in the turning point form. In the turning point form, a 'controls the shape', so a = 3.

So any answer of the form $y = 3(x - h)^2 + 7 \,$ satisfies the given properties. The book chose h = 0, probably the simplest answer. Other concrete answers can be got by choosing h = 1, 2, 3, -1, -2 -3, 1/2, -1/2, ......

so i can pretty much put any value for a... thnx
• Jan 25th 2008, 05:51 PM
mr fantastic
Quote:

Originally Posted by ruscutie100
so i can pretty much put any value for a... thnx

Yes, as long as it's positive.