Question 224743
Complete each of the following for the function g(x)=-3x^2-6x+1


a.Find the vertex of g(x)


b.Find the x-intercept 


c.Find the y-intercept 


{{{g(x)=-3x^2-6x+1}}}


Step 1. The general formula for a parabola is {{{y-b=c(x-a)^2}}} where x=a and y=b is the vertex or at point (a,b).  Here, c is a constant.


Step 2. Put {{{g(x)=y=-3x^2-6x+1}}} in  standard form by completing the square in x with the following steps.


{{{y=-3x^2-6x+1=-3x^2-6x-3+3+1=3(x+1)+4}}}  where we added zero 0=-3+3 to complete the square


Factor out -3 in the first three terms


{{{y=-3(x^2+2x+1)+3+1=-3(x+1)^2+4}}}


Subtract 4 from both sides of the equation



{{{y-4=-3(x+1)^2+4-4}}}


{{{y-4=-3(x+1)^2}}}


ANSWER:  So the vertex is located at point (-1, 4)



Step 3.  To find the x-intercept,  set g(x)=y=0.  We use the quadratic formula given as


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 


where a=-3, b=-6 and c=1


*[invoke quadratic "x", -3, -6, 1 ]


Step 4.  To find the y-intercept set x=0 in g(x)=y=-3x^2-6x+1 


{{{g(0)=y=-3*0^2-6*0+1}}}


or y=1.   


The y-intercept is at (0,1) which is consistent with the above graph.



I hope the above steps and explanation were helpful.


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And good luck in your studies!


Respectfully,
Dr J

http://www.FreedomUniversity.TV