Can I get some help please? Thanks!
2 Answers
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]}
Graph of shape
Thus
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Output
THUS D
Explanation:
Given:
This is the vertex form of a quadratic (completing the square).
If you were to expand the bracket your first term would be
As this is positive the graph is of form
Consider the standardised form of
Then
So we have
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This is one way to remember the relationship
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D
Domain comes before Range
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Output