Question

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?

Answer 1

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.!