Question

Identify the quadratic function that contains the points (-1,-4), (0,0) and (2,-10).

Answer 1

Answer:

Hence, the quadratic equation is:

$f(x)=-3x^2+x=-x(3x-1)$

Step-by-step explanation:

Let the quadratic formula be given by:

$f(x)=ax^2+bx+c$

Now we are given the interpolating points and there corresponding values as:

(-1,-4), (0,0) and (2,-10).

this means then x=-1 f(x)=-4

when x=0 then f(x)=0

and when x=2 then f(x)=-10

so we first put x=0

then f(x)=0=c

hence c=0.

now we are left with the function:

$f(x)=ax^2+bx------(1)$

Hence now we put x=-1 in equation (1)

we get:

$f(x)=a-b=-4------(2)$

now we put x=2

$f(x)=4a+2b=-10------(3)$

On solving equation (2) and (3) by elimination we get:

Multiply equation (2) by 2 and add to equation (3) we obtain;

       2a-2b= -8

       4a+2b= -10

---------------------------

       6a=-18

⇒   a= -3   on dividing both side by 6.

Hence on putting the value of a in equation (2) we get:

b=1

Hence, the quadratic equation is:

$f(x)=-3x^2+x=-x(3x-1)$