First Principles AI

Solution

For $f(x)=ax^2+bx+c$ , the vertex occurs at:

x=h=-\frac{b}{2a}

and the corresponding value is $k=f(h)$ . The vertex is:

(h,k)

Domain of any quadratic polynomial is $\mathbb{R}$ .

Range depends on whether the parabola opens up ( $a>0$ ) or down ( $a<0$ ).

h=-\frac{-4}{2(1)}=2,\qquad k=f(2)=4-8=-4

\boxed{\text{Minimum point }(2,-4)}

\boxed{\text{Domain }\mathbb{R},\;\text{Range }[-4,\infty)}

h=-\frac{-5}{2(1)}=\frac{5}{2}

\begin{aligned} k&=f\left(\frac{5}{2}\right)=\left(\frac{5}{2}\right)^2-5\left(\frac{5}{2}\right)+6\\ &=\frac{25}{4}-\frac{25}{2}+6=\frac{25-50+24}{4}=-\frac{1}{4} \end{aligned}

\boxed{\text{Minimum point }\left(\frac{5}{2},-\frac{1}{4}\right)}

\boxed{\text{Domain }\mathbb{R},\;\text{Range }\left[-\frac{1}{4},\infty\right)}

h=-\frac{2}{2(-1)}=1,\qquad k=f(1)=-1+2-8=-7

\boxed{\text{Maximum point }(1,-7)}

\boxed{\text{Domain }\mathbb{R},\;\text{Range }(-\infty,-7]}

f(x)=(x-2)^2

So the vertex is $(2,0)$ (minimum).

\boxed{\text{Minimum point }(2,0)}

\boxed{\text{Domain }\mathbb{R},\;\text{Range }[0,\infty)}

h=-\frac{2}{2(1)}=-1

k=f(-1)=1-2-8.3=-9.3

\boxed{\text{Minimum point }(-1,-9.3)}

\boxed{\text{Domain }\mathbb{R},\;\text{Range }[-9.3,\infty)}

f(x)=-x^2-x+6

h=-\frac{-1}{2(-1)}=-\frac{1}{2}

\begin{aligned} k&=f\left(-\frac{1}{2}\right)=6-\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)^2\\ &=6+\frac{1}{2}-\frac{1}{4}=\frac{25}{4} \end{aligned}

\boxed{\text{Maximum point }\left(-\frac{1}{2},\frac{25}{4}\right)}

\boxed{\text{Domain }\mathbb{R},\;\text{Range }\left(-\infty,\frac{25}{4}\right]}