Question 971171

Can you help me find the equation of parabola axis vertical
phases through (-1,0), (5,0), (1,8)?
<pre>General form of quadratic equation: {{{ax^2 + bx + c = 0}}}

{{{a(- 1)^2 + b(- 1) + c = 0}}} ------- Using (- 1, 0) in general quadratic equation
{{{a - b + c = 0}}} ------ eq (i)

{{{a(5)^2 + b(5) + c = 0}}} ------- Using (5, 0) in general quadratic equation
{{{25a + 5b + c = 0}}} ------ eq (ii)

{{{a(1)^2 + b(1) + c = 8}}} ------- Using (1, 8) in general quadratic equation
{{{a + b + c = 8}}} ------ eq (iii)

- 2b = - 8 --------- Subtracting eq (iii) from eq (i)
{{{b = (- 8)/(- 2)}}}, or b = 4 

{{{a - 4 + c = 0}}} ------- Substituting 4 for b in eq (i)
a = 4 - c ------- eq (iv)

25(4 - c) + 5(4) + c = 0 ------ Substituting 4 - c for a, and 4 for b in eq (ii)
100 - 25c + 20 + c = 0
- 25c + c + 120 = 0
- 24c = 0 - 120
- 24c = - 120
{{{c = (- 120)/(- 24)}}}, or c = 5 

a = 4 - 5 ------- Substituting 5 for c in eq (iv)
a = - 1

Therefore,
a = - 1
b = 4
c = 5

General quadratic equation: {{{ax^2 + bx + c = 0}}} now becomes:
{{{highlight_green(- x^2 + 4x + 5 = 0)}}}