Respuesta :
EXPLANATION
Vertex:
The vertex of an up-down facing parabola of the form:
[tex]y=ax^2+bx+c[/tex]is:
[tex]x_v=-\frac{b}{2a}[/tex]Rewrite
[tex]y=6(x+3)^2-4[/tex]in the form y=ax^2 +bx + c
Expanding:
[tex]y=6x^2+36x+50[/tex]The parabola params are:
a=6, b=36, c=50
[tex]x_v=-\frac{b}{2a}[/tex][tex]x_v=-\frac{36}{2\cdot6}[/tex]Simplify:
[tex]x_v=-3[/tex]Plug in x_v = -3to find the y_v value
[tex]y_v=6(-3)^2+36(-3)+50[/tex][tex]y_v=-4[/tex]Therefore the parabola vertex is (-3,-4)
a=6 so the vertex is a minimum.
Axis:
Parabola standard equation
4p(y-k) = (x-h)^2 is the standard equation for an up-down facing parabola with vertex at (h,k), and a focal length p
Rewrite y=6(x+3)^2-4 in the standard form:
Add 4 to both sides
y + 4 = 6(x+3)^2 -4+4
Refine
y+4 = 6(x+3)^2
Divide both sides by 6
[tex]\frac{y+4}{6}=\frac{6(x+3)^2}{6}[/tex]Simplify
[tex]\frac{y}{6}+\frac{2}{3}=(x+3)^2_{}_{}[/tex]Factor 1/6
[tex]\frac{1}{6}(y+\frac{\frac{2}{3}}{\frac{1}{6}})=(x+3)^2[/tex]Simplify:
[tex]\frac{1}{6}(y+4)=(x+3)^2[/tex]Factor 4:
[tex]4\cdot\frac{\frac{1}{6}}{4}(y+4)=(x+3)^2[/tex]Simplify:
[tex]4\cdot\frac{1}{24}(y-(-4))=(x-(-3))^2[/tex]Therefore parabola properties are:
[tex](h,k)=(-3,-4),\text{ p=1/24}[/tex]Parabola is of the form 4p(y-k)=(x-h)^2 and is symmetric around the y-axis.
Axis of symmetry is a line parallel to the y-axis wich intersects the vertex:
x = -3
