Question

The minimum value of the polynomial 4x^2-6x+1

Answer 1

Answer:

The minimum value of the polynomial is -1.25.

Step-by-step explanation:

The given polynomial is

$p(x)=4x^2-6x+1$

The minimum value of an upward parabola is its vertex.

The vertex of  $f(x)=ax^2+bx+c$ is

$(\frac{-b}{2a},f(\frac{-b}{2a}))$

$a=4,b=-6,c=1$

$\frac{-b}{2a}=\frac{-(-6)}{2(4)}=\frac{3}{4}=0.75$

Substitute $x=\frac{3}{4}$ in the given equation.

$p(\frac{3}{4})=4(0.75)^2-6(0.75)+1=-1.25$

The vertex of the parabola is (0.75, -1.25). Therefore the minimum value of the polynomial is -1.25.

Answer 2

Answer:

vertex=(0.75,-1.25)

minimum value = -1.25