Question

Find the quadratic function passing through the points (0,-3), (1,2), and (2,-1)

Answer 1

Answer:

The quadratic function passing through the points (0,-3), (1,2), and (2,-1) is $\mathrm{f}(\mathrm{x})=-4 x^{2}+9 x-3$

Solution:

Given that required function is quadratic  

And function is passing through points (0 , -3) , (1 , 2) and (2 , -1)

General form of a quadratic function is  $f(x)=a x^{2}+b x+c$ ----(A)

f(x) is nothing but output value that is y.

That is f(x) = y

So $y=a x^{2}+b x+c$  --- (1)

Let’s use equation (1) to get required function.

Given that function passes through (0 , -3) means when x = 0 , y = -3

On substituting value of x and y in equation (1) we get  

$-3=\mathrm{a}(0)^{2}+\mathrm{b}(0)+\mathrm{c}$

-3 = 0 + 0 + c

c = -3

On substituting value of c in equation (1) we get  

$y=a x^{2}+b x-3$ ---(2)

Function also passes through point (1, 2) that is at x = 1 , y = 2.

On substituting value of x and y in equation (2) we get  

$2=\mathrm{a}(1)^{2}+\mathrm{b}(1)-3$

2 = a + b – 3

a + b = 5                

b = 5 - a   -------(3)

Also given function passes through point ( 2 , -1) means when x = 2 , y = -1

On substituting value of x and y in equation (2) we get  

$-1=\mathrm{a}(2)^{2}+\mathrm{b}(2)-3$

-1 = 4a + 2b – 3

4a + 2b = 2

2a + b = 1  ------- (4)

On substituting value of b from equation (3) in equation (4), we get

2a + (5 - a ) = 1  

a + 5 = 1

a = 1-5 = -4

From equation (3) b = 5 – a = 5 – (-4) = 9

b = 9

Now we have a = -4, b = 9 and c = -3

On substituting calculated values of a, b, and c in equation (A) we get

$f(x)=-4 x^{2}+9 x-3$

Hence required quadratic function is $f(x)=-4 x^{2}+9 x-3$