Question

Write in standard form the quadratic function defined by the points (0, 5), (5, 0), and (3, −4).

Answer 1

Answer:

$f(x)=x^2-6x+5$

Step-by-step explanation:

The general form of quadratic function is

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

It is given that the function defined by the points (0, 5), (5, 0), and (3, −4). It means the function must be satisfied by these points.

For (0,5),

$5=a(0)^2+b(0)+c\Rightarrow c=5$

The value of c is 5.

For (5,0),

$0=a(5)^2+b(5)+(5)\Rightarrow 25a+5b=-5$ .... (2)

For (3,-4),

$-4=a(3)^2+b(3)+(5)\Rightarrow 9a+3b=-9$ .... (3)

Multiply equation (2) by 3 and equation (3) by 5.

$75a+15b=-15$ .... (4)

$45a+15b=-45$ .... (5)

Subtract equation (5) from equation (4).

$75a-45a=-15-(-45)$  

$30a=30$  

$a=1$  

Substitute a=1 in equation (3).

$9(1)+3b=-9$

$9+3b=-9$

$3b=-9-9$

$3b=-18$

$b=-6$

Substitute a=1,b=-6 and c=5 in equation (1).

$f(x)=1x^2+(-6)x+(5)$

$f(x)=x^2-6x+5$

Therefore, the required function is $f(x)=x^2-6x+5$.