The answer or range would be 40
2008
end year using 4 digits
Hey there
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The correct answer is:
f(x+5)=(x+5)^2+3(x+5)-10
f(x+5)=x^2+10x+25+3x+15-10
f(x+5)=x^2+13x+30 and f(x+5)=x^2+kx+30 so
k=13
Now factor x^2+13x+30
Find j and k such that jk=ac=30 and j+k+b=13 so j and k are 10 and 3 so
(x+3)(x+10)
So the two zeros occur when x=-3 and -10 the smallest of which is:
x=-10
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Hope this helps you
