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Determine the coordinates of the vertex for each quadratic function and whether the parabola has a maximum or minimum. 4. y = x² +6x-2​

Sagot :

Fieryanswererftidn

Answer:

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

explanation:

[tex]\sf y =x^2 +6x-2[/tex]

x coordinates on vertex:

solving steps:

  • [tex]\sf \dfrac{-b}{2a}[/tex]
  • [tex]\sf \dfrac{-6}{2(1)}[/tex]
  • [tex]\sf -3[/tex]

Find y-coordinate on vertex:

[tex]\sf y =x^2 +6x-2[/tex]

[tex]\sf y =(-3)^2 +6(-3)-2[/tex]

[tex]\sf y =-11[/tex]

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

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

coordinates: (-3,-11) thus minimum

Semsee45idn

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:

[tex]\dfrac{dy}{dx}=2x+6[/tex]

[tex]\dfrac{dy}{dx}=0\implies 2x+6=0 \implies x=-3[/tex]

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

[tex]\implies (-3)^2+6(-3)-2=-11[/tex]

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

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

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