A) To get the three equations, we will substitute each of the 3 points on the parabola into the quadratic formula
Quadratic function formula is given by:
[tex]y\text{ = }ax^2\text{ + bx + c }[/tex]using point (-1, 5) = (x, y)
[tex]\begin{gathered} 5=a(-1)^2\text{ + b(-1) + c} \\ 5\text{ = a(1) - b + c } \\ 5\text{ = a - b + c }\ldots.(1) \end{gathered}[/tex]using point (0, -4) = (x, y)
[tex]\begin{gathered} -4=a(0)^2\text{ + }b(0)\text{ + c} \\ -4\text{ = c } \end{gathered}[/tex]using point (4, 0)
[tex]\begin{gathered} 0=a(4)^2\text{ + b(4) + c} \\ 0\text{ = 16a + 4b + c} \\ \text{16a + 4b + c = 0 . . . (2)} \end{gathered}[/tex][tex]\begin{gathered} \text{The 3 equations using orderd pair:} \\ EQ1\colon\text{ }5=a(-1)^2\text{ + b(-1) + c} \\ EQ2\colon\text{ }-4=a(0)^2\text{ + b(0) + c} \\ EQ3\colon\text{ }0=a(4)^2\text{ + b(4) + c} \end{gathered}[/tex]B) The linear system:
[tex]\begin{gathered} 5\text{ = a - b + c . . . (1)} \\ -\text{4 = c . . . (2)} \\ \text{0 = 16a + 4b + c . . . (3)} \end{gathered}[/tex]C) substitute for c in equation 1 and 2:
[tex]\begin{gathered} 5\text{ = a - b + c }\ldots.(1) \\ 5\text{ = a - b -4} \\ 5\text{ + 4 = a - b } \\ 9\text{= a - b }\ldots(4) \\ \\ \text{0 = 16a + 4b + c . . . (3)} \\ \text{0 = 16a + 4b }-4 \\ 0+\text{4 = 16a + 4b } \\ 4\text{ = 16a + 4b . . . (5)} \end{gathered}[/tex]Using elimnation for equation (4) and (5):
To eliminate a variable, it must have the same coefficient in both equations.
Let's elimnate b. We will multiply equation (4) by 4 so the coefficient will be the same:
4(9) = 4(a) - b(4)
36 = 4a - 4b ...(4)
4 = 16a + 4b ...(5)
Add equation 4 and 5 together:
36 +4 = 4a + 16a - 4b + 4b
40 = 20a
divide both sides by 20:
40/20 = 20a/20
a = 2
substitute for a in equation 5:
4 = 16(2) + 4b
4 = 32 + 4b
4 - 32 = 4b
-28 = 4b
divide both sides by 4:
-28/4 = 4b/4
b = -7
a = 2, b = -7, c = -4
The equation of the parabola becomes:
[tex]y=2x^2\text{ - 7x - 4}[/tex]