Asked by Alex
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
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
Answers
Answered by
Reiny
#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?
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?
Answered by
Alex
Yesss. Thank you so much!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.