# How do you find the domain and range of y=(x^2) - 6x + 1?

Mar 21, 2017

Domain: All real numbers i.e $\left(- \infty , \infty\right)$
Range: Real numbers $\ge - 8 \mathmr{and} \left[- 8 , + \infty\right)$

#### Explanation:

y=x^2-6x+1 ; a=1 ,b= -6 ; c=1 [ax^2+bx+c]
Domain (possible values of x) : All real numbers i.e $\left(- \infty , \infty\right)$
Range: This is a parabola of which vertex(x) is $- \frac{b}{2} a = \frac{6}{2} = 3$ and Vertex(y) is $y = {3}^{2} - 6 \cdot 3 + 1 = - 8$

So vertex is at $\left(3 , 8\right)$ Since $a = + i v e$. the parabola opens upward and $y = - 8$ is the minimum value and $+ \infty$ is maximum value
So range is $y \ge - 8 \mathmr{and} \left[- 8 , + \infty\right)$ graph{x^2-6x+1 [-20, 20, -10, 10]} [Ans]