Question

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

Answer 1

Answer:

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

explanation:

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

x coordinates on vertex:

solving steps:

• $\sf \dfrac{-b}{2a}$

• $\sf \dfrac{-6}{2(1)}$

• $\sf -3$

Find y-coordinate on vertex:

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

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

$\sf y =-11$

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

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

coordinates: (-3,-11) thus minimum

Answer 2

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.