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 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?
Yesss. Thank you so much!
1) To find the quadratic function that represents the given three points, we can use the method of substitution.
First, let's assume the quadratic function has the form f(x) = ax^2 + bx + c.
Now, let's substitute the x and y values from the first point (0, 6) into the function:
6 = a(0)^2 + b(0) + c
Simplifying this equation gives c = 6.
Next, let's substitute the x and y values from the second point (2, 16) into the function:
16 = a(2)^2 + b(2) + 6
Simplifying this equation gives 4a + 2b = 10.
Similarly, let's substitute the x and y values from the third point (3, 33) into the function:
33 = a(3)^2 + b(3) + 6
Simplifying this equation gives 9a + 3b = 24.
Now, we have a system of two equations:
4a + 2b = 10 (Equation 1)
9a + 3b = 24 (Equation 2)
Solving this system of equations, we find a = 2 and b = -3.
Therefore, the quadratic function that represents the given three points is:
f(x) = 2x^2 - 3x + 6
So, the correct answer is A. f(x)=4x^2−3x+6.
2) Let's follow the same method to find the quadratic function that represents the given three points.
Assume the quadratic function has the form f(x) = ax^2 + bx + c.
Substituting the x and y values from the first point (-1, -8) into the function gives:
-8 = a(-1)^2 + b(-1) + c
Simplifying this equation gives -a - b + c = -8.
Substituting the x and y values from the second point (0, -1) into the function gives:
-1 = a(0)^2 + b(0) + c
Simplifying this equation gives c = -1.
Substituting the x and y values from the third point (1, 2) into the function gives:
2 = a(1)^2 + b(1) - 1
Simplifying this equation gives a + b = 3.
Now, we have a system of two equations:
-a - b + c = -8 (Equation 1)
a + b = 3 (Equation 2)
To eliminate variable c, we substitute c = -1 into Equation 1:
-a - b - 1 = -8
Rearranging this equation gives a + b = 7.
Now, we have the following system of equations:
a + b = 3 (Equation 2)
a + b = 7 (Equation 3)
These equations are inconsistent since they cannot be satisfied simultaneously.
Therefore, there is no quadratic function that passes through the given three points.
3) To find the equation of a parabola that has a vertex (3,5) and passes through the point (1,13), we can use the vertex form of a quadratic equation.
The vertex form of a quadratic equation is given by:
f(x) = a(x - h)^2 + k, where (h, k) represents the coordinates of the vertex.
In this case, the vertex is given as (3,5), so we can substitute h = 3 and k = 5 into the equation:
f(x) = a(x - 3)^2 + 5.
Now, we can substitute the point (1,13) into the equation:
13 = a(1 - 3)^2 + 5
Simplifying this equation gives -6a + 8 = 13,
which further simplifies to -6a = 5.
Solving for a, we find a = -5/6.
Therefore, the equation of the parabola that has a vertex (3,5) and passes through the point (1,13) is given by:
f(x) = (-5/6)(x - 3)^2 + 5.
So, the correct answer is C. y=−2(x−3)2+5.