How do you find the axis of symmetry, and the maximum or minimum value of the function $y = x^{2} - 6 x + 4$ $y=x^{2}-6x+4$ ?

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Answer 1 · 2017-04-04T07:26:48

The minimum value is at $(3, - 5)$ $(3,-5)$

The equation of this parabola is in the form $y = a x^{2} + b x + c$ $y = ax^2 +bx+c$

$y = x^{2} - 6 x + 4 \to a = 1, b = - 6 and c = 4$ $y = x^2 -6x+4" "rarr" "a=1," " b=-6 and c =4$

The axis of symmetry is found from $x = \frac{- b}{2 a}$ $x= (-b)/(2a)$

$x = \frac{- (- 6)}{2 \times 1} = \frac{6}{2} = 3$ $x= (-(-6))/(2xx1) = 6/2=3$

The Turning Point is the maximum or minimum value and it lies on the axis of symmetry, so you have the $x$ $x$ -coordinate

If $x = 3$ $x=3$ , substitute to find $y$ $y$ .

$y = {(3)}^{2} - 6 (3) + 4 = 9 - 18 + 4$ $y = (3)^2 -6(3)+4 = 9-18+4$

$y = - 5$ $y = -5$

In this case the parabola opens upwards ( $a > 0$ $a >0$ , ie positive), therefore the Turning Point is a minimum value.

$(3, - 5)$ $(3,-5)$

How do you find the axis of symmetry, and the maximum or minimum value of the function y=x2−6x+4y=x^{2}-6x+4?