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Sagot :
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
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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