Quadratic function:
y = a x² + b x + c
In this case:
for x = 0 , y = 6
y = a x² + b x + c
6 = a ∙ 0² + b ∙ 0 + c
6 = 0 + 0 + c
6 = c
c = 6
Your quadratic function become:
y = a x² + b x + 6
for x = 2 , y = 16
y = a x² + b x + 6
16 = a ∙ 2² + b ∙ 2 + 6
16 = 4 a + 2 b + 6
Subtract 6 to both sides
10 = 4 a + 2 b
4 a + 2 b = 10
Divide both sides by 2
2 a + b = 5
for x = 3 , y = 33
y = a x² + b x + 6
33 = a ∙ 3² + b ∙ 3 + 6
33 = 9 a + 3 b + 6
Subtract 6 to both sides
27 = 9 a + 3 b
9 a + 3 b = 27
Divide both sides by 3
3 a + b = 9
Now you must solve system:
2 a + b = 5 , 3 a + b = 9
2 a + b = 5
-
3 a + b = 9
_________
- a = - 4
Multiply both sides by - 1
a = 4
Replace this value in equation:
2 ∙ 4 + b = 5
8 + b = 5
Subtract 8 to both sides
b = - 3
So:
a = 4 , b = - 3 , c = 6
y = a x² + b x + c
y = 4 x² - 3 x + 6
Given the following three points, find by hand the quadratic function they represent.
(0,6), (2,16), (3, 33)
- f(x)=−4x2−3x+6
- f(x)=4x2+3x+6
- f(x)=−4x2+21x+6
- f(x)=4x2−3x+6
I am struggling a lot with this one. But i think its A
1 answer