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

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

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Answer 1 · 2017-03-21T12:05:39

Domain: All real numbers i.e $(- \infty, \infty)$ $(-oo,oo)$
Range: Real numbers $\geq - 8 or [- 8, + \infty)$ $>= -8 or [-8, +oo)$

Explanation:

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

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

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

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