Question

Someone, please help with this MATH question.

Answer 1

Answer:

  see attached

Step-by-step explanation:

You have done the hard part: finding the vertex.

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The leading coefficient is the coefficient of the x^2 term. It is +1, a positive number, so the parabola opens upward.

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The x-coordinate of the vertex for a parabola described by ...

  f(x) = ax² +bx +c

is ...

  x = -b/(2a)

For this parabola, a=1, b=6, c=2, and the x-coordinate of the vertex is ...

  x = -6/(2(1)) = -3

The value of the function at that point is ...

  f(-3) = (-3)² +6(-3) +2 = 9 -18 +2 = -7

The coordinates of the vertex are ...

  (-3, -7)

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The axis of symmetry is the vertical line through the vertex. Its equation will be ...

  x = -3

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Some points on the graph can be found by choosing x-values near the vertex value. They will be symmetrical about the axis of symmetry. When the leading coefficient is 1, the y-values will increase above the vertex point by the square of the x-distance from the vertex. The point at x= (-3 +2) will be at y = (-7 +(2²)), or (x, y) = (-1, -3). The symmetrical point is (-5, -3).