Question 1167950
the standard equation of a parabola is ax^2 + bx + c = y


the equation goes through the point (-6,15)


replace x with -6 and y with 15 to get:


a * 36 + b * -6 + c = 15


simplify to get:


36a - 6b + c = 15


the equation also goes through the point (-4,11)


replace x with -4 and y with 11 to get:


a * 16 + b * -4 + c = 11


simplify to get:


16a - 4b + c = 11


you have two equations that need to be solved simultaneously.


they are:


36a - 6b + c = 15
16a - 4b + c = 11


subtract the second equation from the first to get:


20a - 2b = 4


the vertex of the parabola is at (-4,11)


the formula for the vertex in the standard form of the equation is:


x = -b/2a


when x = -4, this equation becomes;


-b/2a = -4


solve for b to get b = 8


the equation of 20a - 2b = 4 becomes:


20a - 16 = 4


add 16 to both sides to get:


20a = 20


solve for a to get:


a = 1


you have:


a = 1 and b = 8


go back to the original equations of:


36a - 6b + c = 15
16a - 4b + c = 11


in either equation, replace a with 1 and b with 8 to get:


36a - 6b + c = 15 becomes 36 - 48 + c = 15 which becomes -14 + c = 15.


solve for c to get c = 27.


you now have a = 1, b = 8 and c = 27


go back to your original equation of ax^2 + bx + c = y and replace a with 1 and b with 8 and c with 17 to get:


x^2 + 8x + 27 = y


graph this equation to get:


when you replace x with -4, the equation becomes:


(-4)^2 + 8*-4 + 27 = y


solve for y to get y = 11


when you replace x with -6, the equation becomes:


(-6)^2 + 8*-6 + 27 = y


solve for y to get y = 15


both points are on the parabola.


the vertex is at x = -b/2a which becomes x = -8/2 which becomes x = -4


the y value was already calculated and is equal to 11 when x = -4.


you can graph the equation to get:


<img src = "http://theo.x10hosting.com/2020/102204.jpg" >


both points are on the parabola and the vertex is at (-4,11).


solution is confirmed to be good.


the standard form of the equaion of the parabols is:


y = x^2 + 3x + 27