SOLUTION: Find an equation in the form y=ax^2+bx+c for the parabola passing through the points (-1,0),(2,-54),(5,-252)

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Find an equation in the form y=ax^2+bx+c for the parabola passing through the points (-1,0),(2,-54),(5,-252)      Log On


   

Question 1206103: Find an equation in the form y=ax^2+bx+c for the parabola passing through the points (-1,0),(2,-54),(5,-252)
Found 4 solutions by MathLover1, greenestamps, ikleyn, Edwin McCravy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

y=ax%5E2%2Bbx%2Bc
given:
(-1,0),(2,-54),(5,-252)

use (-1,0)
0=a%28-1%29%5E2%2Bb%28-1%29%2Bc
a-b%2Bc=0.......eq.1

use (2,-54)
-54=a%282%29%5E2%2Bb%282%29%2Bc
-54=4a%2B2b%2Bc
4a%2B2b%2Bc=-54.......eq.2

use (5,-252)
-252=a%285%29%5E2%2Bb%285%29%2Bc
-252=25a%2B5b%2Bc
25a%2B5b%2Bc=-252.......eq.3

solve the system:
a-b%2Bc=0.......eq.1
4a%2B2b%2Bc=-54.......eq.2
25a%2B5b%2Bc=-252.......eq.3
_________________________
subtract eq.1 from eq.2
4a%2B2b%2Bc-%28a-b%2Bc%29=-54-0
4a%2B2b%2Bc-a%2Bb-c=-54
3a%2B3b=-54
a%2Bb=-54%2F3
a%2Bb=-18
a=-b-18.....eq.1a

go to
a-b%2Bc=0.......eq.1, substitute a
-b-18-b%2Bc=0
-2b%2Bc=18
c=2b%2B18....eq.2a

go to
25a%2B5b%2Bc=-252.......eq.3, substitute a and c
25%28-b-18%29%2B5b%2B2b%2B18=-252
25%28-b-18%29%2B5b%2B2b%2B18=-252
-18b+-+432=-252
252-432=18b
-180=18b
b=-10

go to
a=-b-18.....eq.1a, substitute b
a=-%28-10%29-18
a=10-18
a=-8
go to
c=2b%2B18....eq.2a, substitute b
c=2%28-10%29%2B18
c=-20%2B18
c=-2

your equation is:
y=-8x%5E2-10x-2

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


Plug each (x,y) pair into the general form of the equation to get three equations in a, b, and c; then solve that system of equations.

(-1,0): a-b%2Bc=0 [1]

(2,-54): 4a%2B2b%2Bc=-54 [2]

(5,-252): 25a%2B5b%2Bc=-252 [3]

Eliminate c between [1] and [2], and between [2] and [3], to get two equations in a and b.

[2]-[1]: 3a%2B3b=-54 [4]

[3]-[2]: 21a%2B3b=-198 [5]

Eliminate b between [4] and [5] and solve for a.

[5]-[4]: 18a=-144

a=-144%2F18=-8

Substitute a=-8 in [4] and solve for b:
-24%2B3b=-54
3b=-30
b=-10

Substitute a=-8 and b=-10 in [1] and solve for c:
-8%2B10%2Bc=0
c=-2

ANSWER: y=-8x%5E2-10x-2

Simplified....

y=-2%284x%5E2%2B5x%2B1%29
y=-2%28x%2B1%29%284x%2B1%29

CHECK:
f%28-1%29=%28-2%29%280%29%28-3%29=0
f%282%29=%28-2%29%283%29%289%29=-54
f%285%29=%28-2%29%286%29%2821%29=-252

Answer by ikleyn(52777) About Me  (Show Source):
You can put this solution on YOUR website!
.

The general approach to solving this and similar problems is to form three equations
for coefficients "a", "b" and "c" of the quadratic function, using the three given points,
then solve this system of three equations.


How you will solve the system, it is just another/separate question.


You may use the solver in your calculator;
or you may use some of numerous online calculators in the Internet;
or do it manually using standard substitution or elimination methods.


Sometimes, some simplifications of this general procedure are possible
if the problem has symmetries that facilitate finding the coefficients.

In this given problem, there are no such symmetries,
so the general approach must be used.



Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
(-1,0),(2,-54),(5,-252)
On your TI-84
Press STAT
Press 1
Using the arrow keys and ENTER,
   Enter the 3 given x-coordinates in L1  
   Enter the 3 given corresponding y-coordinates in L2
Press STAT
Press the right arrow once to highlight CALC
Press 5 for QuadReg
Press the down arrow 4 times to highlight CALCULATE
Press ENTER

You'll see the values for a,b, and c on the screen

y=ax2+bx+c
a=-8
b=-10
c=-2
R2=1

The R2=1 tells us that it's a perfect fit.

Edwin