Question 143722
{{{x^2+8x=11}}} Start with the given equation



Take half of the x-coefficient 8 to get 4 (ie {{{(1/2)*8=4}}})


Now square 4 to get 16. (ie {{{4^2=16}}})




{{{x^2+8x+16=11+16}}} Add this result to both sides



{{{x^2+8x+16=27}}} Combine like terms



{{{(x+4)^2=27}}} Factor {{{x^2+8x+16}}} to get {{{(x+4)^2}}}. (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>) 



*[Tex \LARGE x+4=\pm \sqrt{27}] Take the square root of both sides



*[Tex \LARGE x=-4\pm \sqrt{27}] Subtract 4 from both sides



*[Tex \LARGE x=-4\pm 3\sqrt{3}] Simplify {{{sqrt(27)}}} to get {{{3*sqrt(3)}}}. (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)



So the solutions are



*[Tex \LARGE x=-4+3\sqrt{3}] or *[Tex \LARGE x=-4- 3\sqrt{3}]






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a)


{{{x^2+8x=11}}} Start with the given equation




{{{x^2+8x-11=0}}} Subtract 11 from both sides




To find the vertex, we first need to find the axis of symmetry (ie the x-coordinate of the vertex)

To find the axis of symmetry, use this formula:


{{{x=-b/(2a)}}}


From the equation {{{y=x^2+8x-11}}} we can see that a=1 and b=8


{{{x=(-8)/(2*1)}}} Plug in b=8 and a=1



{{{x=(-8)/2}}} Multiply 2 and 1 to get 2




{{{x=-4}}} Reduce



So the axis of symmetry is  {{{x=-4}}}



So the x-coordinate of the vertex is {{{x=-4}}}. Lets plug this into the equation to find the y-coordinate of the vertex.



Lets evaluate {{{f(-4)}}}


{{{f(x)=x^2+8x-11}}} Start with the given polynomial



{{{f(-4)=(-4)^2+8(-4)-11}}} Plug in {{{x=-4}}}



{{{f(-4)=(16)+8(-4)-11}}} Raise -4 to the second power to get 16



{{{f(-4)=(16)+-32-11}}} Multiply 8 by -4 to get -32



{{{f(-4)=-27}}} Now combine like terms



So the vertex is (-4,-27)





b)



From part a) we found the axis of symmetry to be {{{x=-4}}}



c)


Looking at {{{y=x^2+8x-11}}}, we can see that {{{a=1}}}, {{{b=8}}}, and {{{c=-11}}}. Since {{{a>0}}}, this tells us that the parabola opens upward and that there is a minimum. 


To find the minimum, we only need to look at the vertex. Since the vertex is the point (-4,-27), this means that the minimum is {{{y=-27}}}



d) 



Here's a sketch to visually verify our answers



{{{ graph( 500, 500, -11, 10, -29, 10, x^2+8x-11) }}} Graph of {{{y=x^2+8x-11}}}