Find values of a and b so that the quadratic ax^2+bx+c has a local minimum at the point (5,1). What must be true about the relationship between c and the number 1?

Question

Note: Your answer for a and b may involve the variable c.

Answer 1

Solution:

Let the function be y = ax²+bx+c . Find the derivative of the function dy/dx = 2ax+b is the slope of the function at any point. The local minimum occurs at the lowest point of the curve at slope = 0. Set

2ax+b = 0 . At the point ( 5,1), the x coordinate is 5 so substitute 2a(5) + b = 0, 10a + b =0 , Solve for

a = -b/10 , From y = ax² + bx + c set y =1 and x =5, 1 = a(25) + 5b + c , 1 = 25a + 5b + c Substitute

a = -b/10 , 1 = ( -b/10)( 25)+5b + c, 1 = (- 25b /10) + 5b + c , 1 = -2.5b + 5b + c, 1 = 2.5b+c , 2.5b = 1-c,

b = (1-c ) /2.5 , and a = [-(1-c)/2.5] / 10 = (c-1) / 25

if c =1 , then a = (1-1)/ 2.5 = 0 , and b = (1-1)/25 = 0 then in the equation y = ax²+bx+c = (0)(x²) + (0)(x)+1

and y =1 and the equation reduces to a straight line y=1 parallel to the x axis which means at any point x, y is always equal to one