To find the quadratic function [tex]\( y = ax^2 + bx + c \)[/tex] that passes through the points [tex]\( (1, 2) \)[/tex], [tex]\((-3, -26)\)[/tex], and [tex]\((3, -8)\)[/tex], we need to determine the coefficients [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]. Here's the step-by-step process:
1. Set up the system of equations by substituting each point into the function [tex]\( y = ax^2 + bx + c \)[/tex].
- For the point [tex]\( (1, 2) \)[/tex]:
[tex]\[
2 = a(1)^2 + b(1) + c \implies a + b + c = 2
\][/tex]
- For the point [tex]\( (-3, -26) \)[/tex]:
[tex]\[
-26 = a(-3)^2 + b(-3) + c \implies 9a - 3b + c = -26
\][/tex]
- For the point [tex]\( (3, -8) \)[/tex]:
[tex]\[
-8 = a(3)^2 + b(3) + c \implies 9a + 3b + c = -8
\][/tex]
2. Write the system of linear equations derived from the above substitutions:
[tex]\[
\begin{cases}
a + b + c = 2 \\
9a - 3b + c = -26 \\
9a + 3b + c = -8
\end{cases}
\][/tex]
3. Solve this system of equations to find [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex].
- We can solve these equations using any standard method (substitution, elimination, or matrix methods). For simplicity, we use the elimination method:
- Take the first equation:
[tex]\[
a + b + c = 2
\][/tex]
- Now subtract this equation from the second and third equations:
[tex]\[
9a - 3b + c - (a + b + c) = -26 - 2 \implies 8a - 4b = -28 \implies 2a - b = -7 \implies b = 2a + 7
\][/tex]
[tex]\[
9a + 3b + c - (a + b + c) = -8 - 2 \implies 8a + 2b = -10 \implies 4a + b = -5
\][/tex]
- Substitute [tex]\( b = 2a + 7 \)[/tex] into [tex]\( 4a + b = -5 \)[/tex]:
[tex]\[
4a + (2a + 7) = -5 \implies 6a + 7 = -5 \implies 6a = -12 \implies a = -2
\][/tex]
- Substitute [tex]\( a = -2 \)[/tex] back into [tex]\( b = 2a + 7 \)[/tex]:
[tex]\[
b = 2(-2) + 7 = -4 + 7 = 3
\][/tex]
- Substitute [tex]\( a = -2 \)[/tex] and [tex]\( b = 3 \)[/tex] back into [tex]\( a + b + c = 2 \)[/tex]:
[tex]\[
-2 + 3 + c = 2 \implies c = 1
\][/tex]
4. Write the quadratic function using the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[
y = -2x^2 + 3x + 1
\][/tex]
Thus, the function is:
[tex]\[
y = -2x^2 + 3x + 1
\][/tex]
