If f(0) = 0, f(1) = 2, f(2) = 9, and ax^2+bx+c=0, what is the value of f(3)?
+9479 Assuming f(x) = ax2 + bx + c,
f(0) = a(02) + b(0) + c = c and so c = 0
f(1) = a(12) + b(1) + c = a + b + c = a + b + 0 = a + b and so a + b = 2
f(2) = a(22) + b(2) + c = 4a + 2b + c = 4a + 2b + 0 = 4a + 2b and so 4a + 2b = 9
Now since a + b = 2 , we can subtract b from both sides to get a = 2 - b
Now we can substitute 2 - b in for a into the equation 4a + 2b = 9 to get:
4(2 - b) + 2b = 9 distribute 4 to both terms in parenthesees
8 - 4b + 2b = 9 combine like terms
8 - 2b = 9 subtract 8 from both sides of the equation
-2b = 1 divide both sides of the equation by -2
b = -1/2
Now we can substitute -1/2 in for b into the equation a + b = 2 to get:
a + b = 2
a - 1/2 = 2
a = 2 + 1/2
a = 5/2
And so f(x) = \(\frac52\)x2 - \(\frac12\)x + 0
Now to find f(3) , plug in 3 for x into the function.
f(3) = \(\frac52\)(3)2 - \(\frac12\)(3) + 0 = \(\frac{45}2\) - \(\frac32\) = \(\frac{42}{2}\) = 21
