Can I get some help please? Thanks!

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2 Answers
Sep 10, 2017

#(D)"#

Explanation:

#"the equation of a parabola in "color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))#
where (h , k ) are the coordinates of the vertex and a is a multiplier.

#• " if "a>0" then minimum turning point " uuu#

#• " if "a<0" then maximum turning point " nnn#

#"here "a=2>0rArr" minimum turning point"#

#y=2(x+2)^2-3" is in vertex form"#

#rArrcolor(magenta)"vertex "=(-2,-3)#

#2(x+2)^2>=0" for all real x"#

#rArr"minimum value "=-3#

#y" is defined for all real values of x"#

#rArr" domain is " x inRR#

#"range is "y inRR,y>=-3#
graph{2(x+2)^2-3 [-10, 10, -5, 5]}

Sep 10, 2017

Graph of shape #uu#

#"Minimum "->"Vertex"->(x,y)=(-2,-3) #
Thus # y_("minimum")=-3#

Input #color(white)(.)->" Domain "->x->(-oo,+oo) in RR#
Output#->" Range "color(white)("d")->y->[-3,+oo) in RR-> RR>=-3#

THUS D

Explanation:

Given: #y=2(x+2)^2-3#

This is the vertex form of a quadratic (completing the square).

If you were to expand the bracket your first term would be #+2x^2#

As this is positive the graph is of form #uu# thus the vertex is a minimum.

Consider the standardised form of #y=a(x+color(red)(b/(2a)))^2+color(green)(k)#

Then #color(white)(..)x_("vertex")= (-1)xxcolor(red)(b/(2a))color(white)(..)# and #color(white)(..) y_("vertex")=color(green)(k)#

So we have #"Minimum "->"Vertex"->(x,y)=(-2,-3)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This is one way to remember the relationship

Input #color(white)(..)#comes before Output
D #color(white)("dddd.")#comes before #color(white)(.)# R
Domain comes before Range

Input #color(white)(.)->" Domain "->x->(-oo,+oo) in RR#
Output#->" Range "color(white)("d")->y->[-3,+oo) in RR -> -3<=y#

Tony B