Find a polynomial function whose graph passes through (5,13),(10,-11), and (0,3).

Question

y=
(Simplify your answer. Round to three decimal places as needed.)

Answer 1

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

Answer 2

clearly, your function does not pass through (10,-11) since it is positive for all x > 0

Answer 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

Answer 4

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

Answer 5

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