Problems from Polynomial functions: constant, affine and quadratic

Development:

1. The function is affine. (Odd) the degree of the polynomial is $1$. Therefore, $Dom(f)=Im(f)=\mathbb{R}$.

2. The function is constant. Therefore, $Dom(f)=\mathbb{R}$, $Im(f)=-1$.

3. The function is a polynomial of degree $2$. Therefore its domain is $Dom(f)=\mathbb{R}$. To calculate the image first we must look for the vertex:

$$(-{\displaystyle \frac{b}{2a}},-{\displaystyle \frac{b^2-4ac}{4a}})=(-{\displaystyle \frac{4}{-2}},-{\displaystyle \frac{16-4\cdot (-1)\cdot (-1)}{-4}})=(2,3)$$

Since $a<0$, the parabola goes down and therefore, $Im(f)=(-\mathrm{\infty },3]$.

Solution:

1. $Dom(f)=Im(f)=\mathbb{R}$

2. $Dom(f)=\mathbb{R}$, $Im(f)=-1$

3. $Dom(f)=\mathbb{R}$, $Im(f)=(-\mathrm{\infty },3]$, $v=(2,3)$

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