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Marat540 [252]
2 years ago
11

Determine the coordinates of the vertex for each quadratic function and whether the parabola has a maximum or minimum. 4. y = x²

+6x-2​
Mathematics
2 answers:
GREYUIT [131]2 years ago
6 0

Answer:

minimum, coordinates of vertex: (-3,-11)

explanation:

\sf y =x^2 +6x-2

x coordinates on vertex:

<u>solving steps</u>:

  • \sf \dfrac{-b}{2a}
  • \sf \dfrac{-6}{2(1)}
  • \sf -3

<u>Find y-coordinate on vertex:</u>

\sf y =x^2 +6x-2

\sf y =(-3)^2 +6(-3)-2

\sf y =-11

\mathrm{If}\:a < 0,\:\mathrm{then\:the\:vertex\:is\:a\:maximum\:value}

\mathrm{If}\:a > 0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}

coordinates: (-3,-11) thus minimum

stepladder [879]2 years ago
6 0

Answer:

vertex = (-3, -11)

minimum

Step-by-step explanation:

The vertex of a parabola is its turning point (stationary point).

Therefore, the x-coordinate of the vertex can be determined by differentiating the function, setting it zero and solving for x:

\dfrac{dy}{dx}=2x+6

\dfrac{dy}{dx}=0\implies 2x+6=0 \implies x=-3

Substitute found value for x into the original function to find the y-coordinate:

\implies (-3)^2+6(-3)-2=-11

Therefore, the vertex is (-3, -11)

As the leading term of the quadratic function (x^2) is positive, the parabola will open upwards, so the vertex is its minimum point.

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