#1 and #2 are the same type of problem
I will do #1, you do #2 the same way
let ax^2 + bx + c = y be the function ,
for point (0,6) ---- 0 + 0 + c = 6, so c = 6
for point (2,16) --- 4a + 2b + 6 = 16 , 2a + b = 5
for point (3,33) --- 9a + 3b + 6 = 33, 3a + b = 9
subtract those two: a = 4
sub into 2a + b = 5
8 + b = 5, b = 3
then y = 4x^2 + 3b + 6 , which is choice B
for #3, you should know the vertex-yintercept form of a parabola
if the vertex is (3,5), then the equation must be
y = a(x-3)^2 + 5
but (1,13) lies on it, so
13 = a(1-3)^2 + 5
8 = 4a
a = 2
so y = 2(x-3)^2 + 5
so you see that choice?
Please help.!
1) Given the following three points, find by hand the quadratic function they represent. (0,6), (2,16), (3, 33)
A. f(x)=4x2−3x+6
B. f(x)=4x2+3x+6
C. f(x)=−4x2−3x+6
D. f(x)=−4x2+21x+6
2) Given the following three points, find by hand the quadratic function they represent. (−1,−8), (0,−1),(1,2)
A. f(x)=−3x2+10x−1
B. f(x)=−3x2+4x−1
C.f(x)=−2x2+5x−1
D. f(x)=−5x2+8x−1
3) Find the equation of a parabola that has a vertex (3,5) and passes through the point (1,13).
A. y=−3(x−3)2+5
B. y=2(x−3)2+5
C. y=−2(x−3)2+5
D. y=2(x+3)2−5
2 answers
Yesss. Thank you so much!
2 answers