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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option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

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