Answer:
Step-by-step explanation:
Given: f(1)= f(3) = 0
Putting f(1) = a(1)^2 + b(1) + c
= a + b + c
Putting f(3) = a(3)^2 + b(3) + c
= 9a + 3b + c
Putting f(0)=a(0)^2 + b(0) + c = -3
= c = -3
putting c = -3 In equation (1) and equation (2) we get,
a + b - 3 = 0 9a + 3b - 3 = 0
a + b = 3 3(3a+b) = 3
a = 3 - b 3a + b = 1
3(3-b) + b = 1 [Putting a = 3 - b ]
9 - 3b +b = 1
-2b = -8
b = 4
Answer:
b = 4
Step-by-step explanation:
To find the value of b in the quadratic function f(x) = ax² + bx + c, begin by substituting f(0) = -3, and solve for c:
[tex]f(0) = -3\\\\a(0)^2+b(0)+c=-3\\\\0+0+c=-3\\\\c=-3[/tex]
Substitute c = -3 into the function:
[tex]f(x) = ax^2+bx-3[/tex]
Now, substitute f(1) = 0 and f(3) = 0 into the function to form a system of equations:
[tex]f(1)=0\\\\a(1)^2+b(1)-3=0\\\\a+b-3=0[/tex]
[tex]f(3)=0\\\\a(3)^2+b(3)-3=0\\\\9a+3b-3=0[/tex]
So, the system of equations is:
[tex]\begin{cases}a+b-3=0\\9a+3b-3=0\end{cases}[/tex]
Rearrange the first equation to isolate a:
[tex]a+b-3=0\\\\a=3-b[/tex]
Substitute a = 3 - b into the second equation and solve for b:
[tex]9(3-b)+3b-3=0\\\\27-9b+3b-3=0\\\\24-6b=0\\\\6b=24\\\\b=4[/tex]
Therefore, the value of b is:
[tex]\Large\boxed{\boxed{b=4}}[/tex]