Question 118382


{{{x=-8}}} or {{{x=-6}}}  Start with the given roots



{{{x+8=0}}} or {{{x+6=0}}}  Add 8 to both sides for the first equation. Add 6 to both sides for the second equation. 



{{{(x+8)(x+6)=0}}} Now use the zero product property in reverse. Remember the zero product property says that if {{{A*B=0}}} then {{{A=0}}} or {{{B=0}}}



{{{x^2+14x+48=0}}} Now foil. note: Let me know if you need help with foiling.





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Now to find the turning point, we need to find the x-coordinate of the turning point. To do that, simply use this formula: {{{x=-b/2a}}}


So from {{{f(x)=x^2+14x+48}}} we can see that {{{a=1}}} and {{{b=14}}}. 





{{{x=-14/2(1)}}}  Plug {{{a=1}}} and {{{b=-2}}} into the formula. 


{{{x=-14/2}}}  Multiply


{{{x=-7}}} Reduce



So the x-coordinate of the turning point is  {{{x=-7}}} 




Now we have to evaluate {{{f(-7)}}} to determine the y-coordinate of the turning point.




{{{f(x)=x^2+14x+48}}} Start with the given function.



{{{f(-7)=(-7)^2+14(-7)+48}}} Plug in {{{x=-7}}}. In other words, replace each x with -7.



{{{f(-7)=(49)+14(-7)+48}}} Evaluate {{{(-7)^2}}} to get 49.



{{{f(-7)=(49)+-98+48}}} Multiply 14 and -7 to get  -98



{{{f(-7)=-1}}} Now combine like terms



So the turning point is at (-7,-1)





Notice if we graph {{{f(x)=x^2+14x+48}}}, we can see that the roots are{{{x=-8}}} and {{{x=-6}}}. We can also see that the turning point is (-7,-1). So this visually verifies our answer.




{{{ graph(500,500,-10,10,-10,10,0, x^2+14x+48) }}}