Question

I need help with setting up/starting these problems.
1) y=ax^2+bx+c; vertex (1,4), line passes through (-1,-8). find a b & c.
answers:a=-3, b=6, c=1
2) find the point on y=x-1 that is close to (4,1)
answer: (2,3)

Answer 1

x coordinate for vertex : -b/2a
y coordinate (you plug it into the y)
so -b/2a = 1
-b = 2a
now your parabola will look like :
y = ax^2 -2ax + c
now if you put x=1 y should be 4
4 = a -2a + c
c = 4 + a
so your parabola :
y = ax^2 -2ax + 4 + a
since the parabola passes through -1 , -8
-8 = a + 2a +4 + a
4a = -12
a = -3
b= -2a = 6
c = 4+a = 1

2) distance from the line to the point : ((x-4)^2 + (x-2)^2)^0.5
derivative will be : (4x-12)/((2x^2 -12x +2x)^0.5)
equate derivative to zero : 4x-12 = 0 x = 3
y = x-1 y(3) = 2
(3,2)