How to find the equation of the parablola with 3 points of it known (1 Viewer)

[Noobtron]

Well the title explains everything,

would i sub them into the general parabola equation then do sim equations?

help appreciated

 

[Mattamz]

let the parabola be y=ax^2 + bx + c
then sub in each of the points, 3 equations 3unknowns simulatious equations, to find a, b and c.

do u have a specific question? if post it up

 

[Captain Gh3y]

To elaborate, a parabola is of the form P(x) = ax²+bx+c, and 3 points define a parabola (like how 2 points define a line).

You're given 3 points, say (x₁, P(x₁)), (x₂, P(x₂)), and (x₃, P(x₃)). We want to find the coefficients a, b, and the constant c that give us a parabola through the three points.

Substitute these values of x and P(x) into the general equation of a Parabola so you have

x₁²a + x₂b + c = P(x1)
x₂²a + x₂b + c = P(x2)
x₃²a + x₃ b+ c = P(x3)
remembering the x's are all just numbers and a, b, c are the unknowns

You now have 3 unknowns and 3 equations so you can (usually) solve for all 3.
A fast way is to add or subtract the equations from each other to eliminate some unknowns.

Example: Find the parabola through (1, 13), (3, 30) and (-1, 10)

Our equations for the example become

a + b + c = 13 ........... (1)
9a + 3b + c = 30 ........(2)
a - b + c = 10 .............(3)

So you can see subtracting equation (3) from equation (1) we get
2b = 3
b = 1.5

So now we can use
a + 1.5 + c = 13 and 9a + 3(1.5) + c = 30, which simplify to
a + c = 11.5 ........(4)
9a + c = 25.5 .......(5)

by subtracting equation equation (4) from equation (5) we get
8a = 14
a = 1.75

So now we can use
1.75 + 1.5 + c = 13 which simplifies to
c = 9.75

So the parabola is y = 1.75x² + 1.5x + 9.75

Hope that makes sense

 

[Noobtron]

find the equation of a quadratic function that passes through the points ( -2, 18) (3, -2) and (1,0)

 

[Noobtron]

heres another question,

Find the value for n for which the equation (n+2)x^2 + 3x -5 = 0 has one root triple the other

 

[watatank]

heres another question,

Find the value for n for which the equation (n+2)x^2 + 3x -5 = 0 has one root triple the other

i *think* this is how you do it. i might be wrong though, feel free to correct.

roots A, 3A

sum of roots = A + 3A = 4A = -b/a = -3/(n+2)

product of roots = 3A² = c/a = -5/(n+2)

you've now got two simultaneous equations

i) 4A = -3/(n+2)

ii) 3A² = -5/(n+2)

consider i)
4A = -3/(n+2)

.'. A = -3/4(n+2)

sub into ii)

3A² = -5/(n+2)

3* [ -3 / 4(n+2) ]² = -5/(n+2)

3* [ 9 / 16(n+2)² ] = -5/(n+2)

[ 27 / 16(n+2)² ] = -5/(n+2)


[ 27 / 16(n+2) ] = -5 .................. (n =/ -2)

27 = -5 * 16(n+2)

27 = -80n - 160

187 = -80n

n = -2.3375

 

[Noobtron]

i *think* this is how you do it. i might be wrong though, feel free to correct.

roots A, 3A

sum of roots = A + 3A = 4A = -b/a = -3/(n+2)

product of roots = 3A² = c/a = -5/(n+2)

you've now got two simultaneous equations

i) 4A = -3/(n+2)

ii) 3A² = -5/(n+2)

consider i)
4A = -3/(n+2)

.'. A = -3/4(n+2)

sub into ii)

3A² = -5/(n+2)

3* [ -3 / 4(n+2) ]² = -5/(n+2)

3* [ 9 / 16(n+2)² ] = -5/(n+2)

[ 27 / 16(n+2)² ] = -5/(n+2)


[ 27 / 16(n+2) ] = -5 .................. (n =/ -2)

27 = -5 * 16(n+2)

27 = -80n - 160

187 = -80n

n = -2.3375

yes you were right, watatank

but with the method for the first question posted,

i only got the three simultaneous equations, how can this be solved

plus isnt this line
x1²a + x2b + c = P(x1)

x1²a + x1b + c = P(x1)
?

 

[Captain Gh3y]

They aren't lines, they're 3 linear equations with 3 unknowns a, b & c, we can solve them using the same techniques from simultaneous equations you'd normally use. The x's are the values given to you in the question.

find the equation of a quadratic function that passes through the points ( -2, 18) (3, -2) and (1,0)

So here when you substitute the 3 given points into the general equation of a parabola, y = ax² + bx +c, you get

a(-2)² + b(-2) + c = 18

a(3)² + b(3) + c = -2

a (1)² + b(1) +c = 0

or
4a - 2b + c = 18
9a + 3b + c = -2
a + b + c = 0

So those are the 3 simultaneous equations which will allow us to solve for the unkowns a, b, and c which define the parabola, y = ax² + bx + c.

You can use substitution, elimination or whatever you like to do to solve the 3 simultaneous equations for a, b and c

It works out as
a = 1, b = -5, c = 4

So the parabola is y = (1)x² + (-5)x + (4)
or y = x² - 5x + 4

 

[Noobtron]

k thanks, im a not very good at simultaneous equations

 

[Noobtron]

heres another challenging question

find the values of p for which x^2 - x + 3p -2 > 0 for all x

and another

solve 2^ ( 2x+1) - 5.2 ^ (x) + 2 = 0

 

[Mattamz]

heres another challenging question

find the values of p for which x^2 - x + 3p -2 > 0 for all x

and another

solve 2^ ( 2x+1) - 5.2 ^ (x) + 2 = 0

f(x) = x^2 -x +3p-2
f'(x) = 2x-1 = 0, when x=1/2
f(1/2) = 1/4 - 1/2 +3p-2
= 3p - 2.25

3p-2.25>0
3p>2.25
therefore: p>0.75

hope thats the answer

 

[Mattamz]

heres another challenging question

find the values of p for which x^2 - x + 3p -2 > 0 for all x

and another

solve 2^ ( 2x+1) - 5.2 ^ (x) + 2 = 0

2^ ( 2x+1) - 5.2 ^ (x) + 2 = 0
2.[2^(x)]^2 - 5.2^x +2 = 0
let 2^x = u
2u^2 - 5u + 2 = 0
(2u - 1)(u - 2) = 0
u = 1/2 or u = 2
2^x = 1/2 2^x = 2
x= -1 x=1

therefore x= -1, 1

 

[Noobtron]

heres another

express (4x+1) / ( x^2 - x - 2) in the form


a / (x-2) + b\ (x+1)

 

[watatank]

heres another

express (4x+1) / ( x^2 - x - 2) in the form


a / (x-2) + b / (x+1)

factorise denominator...

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

so now youre trying to put

(4x+1) / (x+1)(x-2)

in the form

a / (x-2) + b / (x+1)


(4x+1) / (x+1)(x-2) = a / (x-2) + b / (x+1)

make RHS hav a common denominator..

(4x+1) / (x+1)(x-2) = [ a(x+1) + b(x-2) ] / [ (x-2)(x+1) ]

denominators are the same, so numerators are same.

(4x+1) = a(x+1) + b(x-2)

(4x+1) = ax+a + bx -2b

(4x+1) = x(a+b) + a -2b

since both sides have to be the same

a + b = 4
a -2b = 1

solve simultaneously.

3b = 3.... b =1

.'. a = 3

(4x+1) / (x+1)(x-2) = 3 / (x-2) + 1 / (x+1)

 

[Noobtron]

how do we do these

x^2 + 5^x - 3 =(triple =) ax(x + 1) + b(x+1) ^2 + cx

 

[Noobtron]

find the exact values for k for which x^2 + 2kx + k + 5 = 0 has real roots

 

[watatank]

find the exact values for k for which x^2 + 2kx + k + 5 = 0 has real roots

are we doing your homework for you?

use the discriminant. if you remember the quadratic formula its the thing in the square root.

discriminant = b^2 - 4ac

if discriminant > 0 it has two roots

discriminant = 0 then it has one root

discriminant < 0 it has no roots

youre looking for (2k)^2 - 4(1)(k+5) > 0

4k^2 - 4k - 20 > 0

4(k^2 - k - 5) > 0

use quadratic formula to solve for k.

 

[Noobtron]

lol im just lost with our teacher going through a whole topic in 2-3 periods and theres no time to sink in information as we have to copy down notes, ive then been handed these hard questions and ive had no idea, but this helps really helped me understand the topic better