Question

How to answer these two questions ?

Answer 1

`f(x)=2x^2-8x-5.`

Explanation:

`f(x)=y=ax^2+bx+c,` intersects the `Y-` Axis in the point

`(0,-5).`

`:. f(0)=-5 rArr c=-5...................(1).`

We know that, for `f_min, f'(x)=0, and, f''(x) gt 0.`

Now, `f(x)=ax^2+bx+c rArr f'(x)=2ax+b, and, f''(x)=2a.`

Since, `f''(x) gt 0, 2a >0, or, a >0.`

`:. f'(x)=0 rArr 2ax+b=0 rArr x=-b/(2a).`

`:. f_min=f(-b/(2a))=a(-b/(2a))^2+b(-b/(2a))+c,`

`=b^2/(4a)-b^2/(2a)+c=c-b^2/(4a)=-5-b^2/(4a).`

This means that `(-b/(2a), -5-b^2/(4a))` is the minimum point of `f.`

`:. (-b/(2a), -5-b^2/(4a))=(2,-13).`

`:. -b/(2a)=2...(2), &, -5-b^2/(4a)=-13 rArr -b^2/(4a)=-8,`

`i.e., b/2(-b/(2a))=-8 :. (2) rArr b/2(2)=-8, or, b=-8.`

Then, by `(2), -(-8)/(2a)=2 rArr a=2; &, a>0," as reqd."`

Altogether, with `a=2, b=-8, and, c=-5,` we get,

`f(x)=2x^2-8x-5,` as desired.

Enjoy Maths.!