The minimum point of a quadratic curve is (1, −4). The curve cuts the y-axis at −1. Show that the equation of the curve is y = 3x2 − 6x −1?

1 Answer
May 29, 2018

Kindly refer to Explanation.

Explanation:

Suppose that, the desired eqn. of the quadratic curve C is

# C : y=f(x)=ax^2+bx+c, (a!=0)#.

For #C nn Y"-Axis"={(0,-1)}. :. f(0)=-1#.

#:. a(0)^2+b(0)+c=-1, i.e., c=-1#.

# :. C : y=f(x)=ax^2+bx-1#.

For #f_min, f'(x)=0 and f''(x) gt 0#.

But, we are given that #f_min=-4=f(1)#.

# :. f'(1)=0, and, f''(1) gt 0#.

#because, f'(x)=2ax+b and f''(x)=2a#, we have,

# f'(1)=0 rArr 2a+b=0 and f''(1) gt 0 rArr a gt 0#.

Also, #f(1)=-4 rArr a+b-1=-4, or, a+b=-3#.

Solving

#2a+b=0 and a+b=-3," we get, "a=3, b=-6#.

Altogether, # C : y=f(x)=3x^2-6x-1#, as desired!

Enjoy Maths.!