Asked by David
I've been staring at this problem for an hour and have NO IDEA WHAT TO DO!
"A quadratic function in the form f(x)=ax^2+bx+c contains the points (-1,0), (0,2), and (2,0)."
a) determine the values of a,b, and, c
b) Factorise the expression ax^2+bx+c obtained from the function in part (a)
please help! i've tried EVERYTHING :(
"A quadratic function in the form f(x)=ax^2+bx+c contains the points (-1,0), (0,2), and (2,0)."
a) determine the values of a,b, and, c
b) Factorise the expression ax^2+bx+c obtained from the function in part (a)
please help! i've tried EVERYTHING :(
Answers
Answered by
MathMate
Let
f(x)=ax^2+bx+c
from (-1,0), we conclude that
f(-1)=0, or
a(-1)^2 + b(-1) + c = 0......(1)
Similarly, from (0,2), we have
a(0)^2 + b(0) + c = 2........(2)
and from (2,0), we get
a(2)^2 + b(2) + c = 0........(3)
By solving the system of linear equations (1),(2) and (3), you will find the values of a,b and c.
After substituting the numerical values of a,b and c into f(x), you can factorize as required in question (c).
Note: the system of linear equations is easy to solve because if you examine equation (2), it reduces to c=2.
By substituting the value of c in (1) and (3), you only have to solve a system of 2 equations with two unknowns (a and b).
f(x)=ax^2+bx+c
from (-1,0), we conclude that
f(-1)=0, or
a(-1)^2 + b(-1) + c = 0......(1)
Similarly, from (0,2), we have
a(0)^2 + b(0) + c = 2........(2)
and from (2,0), we get
a(2)^2 + b(2) + c = 0........(3)
By solving the system of linear equations (1),(2) and (3), you will find the values of a,b and c.
After substituting the numerical values of a,b and c into f(x), you can factorize as required in question (c).
Note: the system of linear equations is easy to solve because if you examine equation (2), it reduces to c=2.
By substituting the value of c in (1) and (3), you only have to solve a system of 2 equations with two unknowns (a and b).
Answered by
Damon
A quadratic function in the form f(x)=ax^2+bx+c contains the points (-1,0), (0,2), and (2,0)."
Another way
y = (x+1)(x-2)(k)
2 = (1)(-2)(k)
2 = -2k
k = -1
so
y = (x+1)(x-2)(-1) (factored already)
so
y = -x^2 + x + 2
y =
Another way
y = (x+1)(x-2)(k)
2 = (1)(-2)(k)
2 = -2k
k = -1
so
y = (x+1)(x-2)(-1) (factored already)
so
y = -x^2 + x + 2
y =
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