Find the quadratic function y=ax+bc+c passing through (-1,-13),(1,-3),(-3,-15)?

The quadratic function should be y = ax² + bx + c instead.

The quadratic function passes through (-1, -13) :
-13 = a(-1)² + b(-1) + c
a - b + c = -13 ...... [1]

The quadratic function passes through (1, -3) :
-3 = a(1)² + b(1) + c
a + b + c = -3 ...... [2]

The quadratic function passes through (-3. -15:
-15 = a(-3)² + b(-3) + c
9a - 3b + c = -15 ...... [3]

[2] - [1] :
2b = 10
b = 5

[2] - [3] :
-8a + 4b = 12
-2a + b = 3 ...... [4]

Substitute b = 5 into [4] :
-2a + 5 = 3
-2a = -2
a = 1

Substitute a = 1 and b = 5 into [2] :
1 + 5 + c = -3
c = -9

The quadratic function is : y = x² + 5x - 9

Do you mean y = ax² + bx + c ?

Plug each point into the equation to get a system of three equations:
(-1,-13):
-13 = a - b + c

(1,-3):
-3 = a + b + c

(-3,-15):
-15 = 9a - 3b + c

Solve the system for a, b, and c.
(a,b,c) = (1,5,-9)

y = x² + 5x - 9

Do you mean y = ax² + bx + c?

Solve the following system for a, b, and c

(1) -13 = a - b + c
(2) -3 = a + b + c
(3) -15 = 9a - 3b + c

-13 - -3 = -2b (1) - (2)
b = 5

(3) 0 = 9a + c
(1) -3 = a + c
_______________
8a = 3
a = 3/8, b = -27/8

Probably a mistake there, but you get the idea

y=ax + bc + c passing through (-1, -13), (1, -3), (-3, -15)

Substitute: x = -1, y = -13

-13 = -a + bc + c

Substitute: x = 1, y = -3

So: -3 = a + bc + c

Substitute: x = -3, y = -15

So: -15 = -3a + bc + c

So: bc + c = -3 + a
and: bc + c = -15 + 3a

So: -15 + 3a = -3 - a

Therefore: a = 3.

Therefore: bc + c = -13 + a = -13 + 3 = -10

So: c(b+1) = -10
(b+1) = -10/c

Therefore: c(b+1) = -15 + 3 x 3 = -15 + 9 = -6
So: c = -6 / (b+1) = -6 x c/-10 = 6c / 10 = 3c/5

So: c = 3/5c
c - 3/5c = 0
c(1-3/5)=0
Therefore: c=0.

Therefore: -13 = -3 + b(0) + 0
So: -10 = b x 0
Therefore: b = -10/0 = -undefined.

Substitute the values of a, b, and c into the quadratic malfunction:

y = 3x - undefined x 0 + 0
Therefore:
y = 3x.

A(-1,-13), B(1.-3), and C(-3,-15) are all on y = ax^2 + bx + c.
A on y--> a - b + c = - 13..(1).
B on y--> a +b + c = - 3..(2).
C on y--> 9a -3b +c = -15..(3).
((2)-(1))--> b = 5..(4).
((3)+3(2))--> 12a + 4c = -24, ie., 3a + c = -6..(5).
Substitution of 5 for b in (2) gives a+ c = -8..(6).
((5)-(6))--> 2a = 2, ie., a = 1. Then (2)--> c = -(a+b+3)= -(1+5+3) = -9 &
y = x^2 + 5x - 9.

yes

Ok

y(-1) = a - b + c
y(1) = a + b + c
y(-3) = a - 3b + 9c

So here is what you have now:

a - b + c = -13
a + b + c = -3
a - 3b + 9c = -15

[1 -1 1]
[1 1 1] = A
[1 -3 9]

[a]
[b] = X
[c]

[-13]
[-3 ] = B
[-15]

Now multiply the inverse of matrix A by B to solve for matrix X:

(1/A)B = X

a = -9
b = 5
c = 1

So your quadratic equation should be:
y = x^2 + 5x - 9