Complete the square to rewrite y = x^2 – 6x + 2 in vertex form. then state whether the vertex is a maximum or a minimum and give its coordinates. a. maximum at (–3, –7) b. minimum at (–3, –7) c. minimum at (3, –7) d. maximum at (3, –7)

Vertex of the equation will be denoted by the point given in option (C).

  Quadratic equation given in the question is,

To convert this equation into vertex form we will rewrite the equation into square form.

y = x² - 6x + 2

  = x² - 2(3x) + 2

  = x² - 2(3x) + (3)² - (3)² + 2

  = [x² - 2(3x) + 3²] - 3² + 2

  = (x - 3)² - 9 + 2

y = (x - 3)² - 7

By comparing this equation with the vertex form of the quadratic equation,

y = a(x - h)² + k

Vertex of the parabola will be (3, - 7).

And the leading coefficient 'a' = 1

Since, leading coefficient is positive, parabola will open upwards.

Therefore, vertex will be the minimum or lowest point of the curve.

     Hence, Option (C) will be the answer.

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i used to know how to do this in 7th but i forget easily

step-by-step explanation:

we have three similar triangles, because each has a right angle and shares an angle.   let's write the angles in order: opposite to short leg, long leg, hypotenuse.

cab similar to bad similar to cbd

or as ratios,

ca: ab: cb = ba: ad: bd = cb: bd: cd

we also know

ac = ad + cd

(ad+cd): ab: cb = ba: ad: bd = cb: bd: cd

(ad+cd)/cb=cb/cd

we have cb=6, ad=5 and seek x=cd.

we reject the negative root and conclude x=4

answer: 4