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Could somebody explain how to make the equation of parabola whose lower point A is (-5;2) and there is also another point B which is (0;27).
General equation is y=ax^2+bx+c. Because the point B cut the y-axis the letter c must be 27 but what are letters a and b ?
Given, two points on the parabola: (0,27) and (-5,2).
If (-5,2) is a lower point on the parabola, then there are infinitely many parabolas that pass through those two points.
If (-5,2) is the lowest point on the parabola, then there is only one parabola that passes through those two points.
I assume you mean (-5,2) is the lowest point on the parabola.
(-5,2) then is the vertex of the vertical parabola that opens upward.
At the vertex, the slope of the tangent line is zero. So, the 1st derivative at that vertex is equal to zero.
y = ax^2 +bx +c ----(i)
y' = 2ax +b
0 = 2a(-5) +b ----at the vertex, x = -5 and y'=0.
0 = -10a +b
b = 10a -----(ii)
You got c=27, which is okay.
Substitute that, and (-5,2), into (i),
2 = a(-5)^2 +b(-5) +27
2 = 25a -5b +27
Isolate b,
5b = 25a +27 -2
5b = 25a +25
b = 5a +5 ------(iii)
b from (ii) = b from (iii),
10a = 5a +5
10a -5a = 5
5a = 5
a = 1 ----***
Plug that in, say, (ii),
b = 10a = 10(1) = 10 ----***
Therefore, the parabola in question is
y = ax^2 +bx +c
y = (1)x^2 +(10)x +27
y = x^2 +10x +27 -----------answer.
Thank you for the detailed explanation