Find a polynomial function whose graph passes through (5,13),(10,-11), and (0,3).
y=
(Simplify your answer. Round to three decimal places as needed.)
clearly, your function does not pass through (10,-11) since it is positive for all x > 0
still wrong, since f(10) = 17
Let's get it right this time
c=3, so we have
25a + 5b + 3 = 13
100a + 10b + 3 = -11
which yields
a = -17/25
b = 27/5
giving
f(x) = -0.68x^2 + 5.4x + 3
Apologies for the confusion in the previous responses. Thank you for pointing out the mistake. After rechecking the calculations, the correct polynomial function that passes through (5,13), (10,-11), and (0,3) is:
f(x) = -0.68x^2 + 5.4x + 3
To find the polynomial function, we use the general form of a polynomial of degree 2:
y = ax^2 + bx + c
We can substitute the x and y values from each of the given points to get a system of equations:
(5, 13) -> 13 = 25a + 5b + c
(10, -11) -> -11 = 100a + 10b + c
(0, 3) -> 3 = 0a + 0b + c
The third equation implies that c = 3.
Now we can substitute the value of c in the first two equations:
13 = 25a + 5b + 3 -> 10 = 25a + 5b
-11 = 100a + 10b + 3 -> -14 = 100a + 10b
Simplifying these equations gives us the system:
10 = 25a + 5b
-14 = 100a + 10b
We can solve this system using elimination or substitution. Let's use elimination:
Multiply the first equation by 2 and the second equation by 5:
20 = 50a + 10b
-70 = 500a + 50b
Now subtract the second equation from the first:
90 = 450a
a = 90/450
a = 0.2
Substitute the value of a in one of the equations to solve for b:
10 = 25(0.2) + 5b
10 = 5 + 5b
5 = 5b
b = 1
Now we have the values of a, b, and c:
a = 0.2
b = 1
c = 3
Therefore, the polynomial function that passes through (5,13), (10,-11), and (0,3) is:
y = 0.2x^2 + x + 3
I apologize for the mistake in my previous response. Let's find the correct polynomial function using the given points:
We can use the general form of a polynomial of degree 2:
y = ax^2 + bx + c
Substituting the given points into the equation, we get the following system of equations:
(5, 13): 13 = 25a + 5b + c
(10, -11): -11 = 100a + 10b + c
(0, 3): 3 = c
Therefore, we have the following equations:
25a + 5b + c = 13
100a + 10b + c = -11
c = 3
Substituting c = 3 into the first two equations:
25a + 5b + 3 = 13
100a + 10b + 3 = -11
Simplifying these equations gives us:
25a + 5b = 10
100a + 10b = -14
To eliminate b, we can multiply the first equation by 2 and subtract it from the second equation:
200a + 20b = -28
-50a - 10b = -10
Simplifying this equation:
150a = -18
a = -18/150
a = -0.12
Substituting the value of a into the first equation to solve for b:
25(-0.12) + 5b = 10
-3 + 5b = 10
5b = 13
b = 13/5
Therefore, the polynomial function that passes through (5,13), (10,-11), and (0,3) is:
y = -0.12x^2 + (13/5)x + 3