How do you find the VERTEX of a parabola $y = x^{2} - x - 6$ $y=x^2-x-6$ ?

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How do you find the VERTEX of a parabola $y = x^{2} - x - 6$ $y=x^2-x-6$ ?

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Answer 1 · 2015-07-21T21:56:12

The vertex is $(\frac{1}{2} - \frac{25}{4})$ $(1/2-25/4)$ or $(\frac{1}{2}, - 6 \frac{1}{4})$ $(1/2, -6 1/4)$ .

Explanation:

$y = x^{2} - x - 6$ $y=x^2-x-6$ is a quadratic equation in the form of $a x^{2} + b^{2} + 6$ $ax^2+b^2+6$ , where $a = 1, b = - 1, and c = - 6$ $a=1, b=-1, and c=-6$ .

The vertex is the minimum or maximum point of the equation. The $x$ $x$ value can be found using the formula $x = \frac{- b}{2 a}$ $x=(-b)/(2a)$ . The $y$ $y$ value can be found by substituting the value for $x$ $x$ into the equation and solving for $y$ $y$ .

The value of $x$ $x$ .

$x = \frac{- b}{2 a} = \frac{- (- 1)}{2 \cdot 1} = \frac{1}{2}$ $x=(-b)/(2a)=(-(-1))/(2*1)=1/2$

$x = \frac{1}{2}$ $x=1/2$

The value of $y$ $y$ .

$y = {(\frac{1}{2})}^{2} - 1 (\frac{1}{2}) - 6$ $y=(1/2)^2-1(1/2)-6$ =

$y = \frac{1}{4} - \frac{1}{2} - 6$ $y=1/4-1/2-6$ =

The common denominator for the right side is $4$ $4$ . Multiply each term by a fraction so that it has a denominator of $4$ $4$ .

$y = \frac{1}{4} - \frac{1}{2} (\frac{2}{2}) - 6 (\frac{4}{4})$ $y=1/4-1/2(2/2)-6(4/4)$ =

$y = \frac{1}{4} - \frac{2}{4} - \frac{24}{4}$ $y=1/4-2/4-24/4$ =

$y = - \frac{25}{4}$ $y=-25/4$

Simplify the improper fraction.

$y = - 6 \frac{1}{4}$ $y=-6 1/4$

graph{y=x^2-x-6 [-14.24, 14.23, -7.12, 7.12]}

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How do you find the VERTEX of a parabola $y = x^{2} - x - 6$ $y=x^2-x-6$ ?

1 Answer

Explanation:

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Science

Math

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How do you find the VERTEX of a parabola y=x2−x−6y=x^2-x-6?

1 Answer

Explanation:

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How do you find the VERTEX of a parabola $y = x^{2} - x - 6$ $y=x^2-x-6$ ?