If you can helpme on this problem you are GOD.
If f(x) = ax^2 + bx + c, what can you say about the values of a, b, and c if:
(A) (1,1) is on the graph of f(x)?
(B) (1,1) is the vertex of the graph of f(x)?[Hint: The axis of symmetry is x = -b/(2a).]
(c)The y intercept of the graph is (0,6)?
(D) Find a quadriatic function satisfying all three conditions.
from (A) we would know that 1 = a + b + c
from (b) we know 1 = -b((2a)
or b = -2a
from (c) we know that at the y-intercept x = 0
so 6 = 0 + 0 + c, so c=6
so back in the first equation
a+b+c = 1
a + (-2a) + 6 = 1
-a = -5
a + 5, then b = -10 and c=6
so f(x) = 5x^2 - 10x + 6
To determine the values of a, b, and c based on the given conditions, let's go step by step.
(A) If (1,1) is on the graph of f(x), it means that when x = 1, f(x) = 1. We can substitute these values into the quadratic equation f(x) = ax^2 + bx + c and get:
1 = a(1)^2 + b(1) + c
1 = a + b + c
Therefore, we can say that a + b + c = 1.
(B) If (1,1) is the vertex of the graph of f(x), it means that the axis of symmetry of the graph is x = 1. According to the given hint, the axis of symmetry is given by x = -b/(2a). Setting x = 1, we can solve for b:
1 = -b/(2a)
2a = -b
b = -2a
(C) If the y-intercept of the graph is (0,6), it means that when x = 0, f(x) = 6. Substituting these values into the quadratic equation, we get:
6 = a(0)^2 + b(0) + c
6 = c
Therefore, we can say that c = 6.
(D) Now, let's find the quadratic function satisfying all three conditions. We know that a + b + c = 1, b = -2a, and c = 6. Substituting these values into the quadratic equation, we have:
f(x) = ax^2 + bx + c
f(x) = ax^2 + (-2a)x + 6
f(x) = ax^2 - 2ax + 6
Therefore, the quadratic function satisfying all three conditions is f(x) = ax^2 - 2ax + 6, where a can be any real number.