Question 118378
To find the roots of {{{f(x)=x^2-4x-5}}}, set the entire function equal to zero


 {{{x^2-4x-5=0}}} Set the function equal to zero



{{{(x-5)(x+1)=0}}} Factor the left side (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>)




Now set each factor equal to zero:

{{{x-5=0}}} or  {{{x+1=0}}} 


{{{x=5}}} or  {{{x=-1}}}    Now solve for x in each case



So the roots are:

 {{{x=5}}} or  {{{x=-1}}} 


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Now let's find the vertex (ie the turning point)


To find the x coordinate of the vertex, simply use this formula {{{x=-b/2a}}}


Looking at {{{f(x)=x^2-4x-5}}}, we can see that {{{a=1}}} and {{{b=-4}}}. 





{{{x=-(-4)/2(1)}}}  Plug {{{a=1}}} and {{{b=6}}} into the formula. 


{{{x=-(-4)/2}}}  Multiply


{{{x=4/2}}}  Negate -4 to get 4


{{{x=2}}} Reduce



So the x-coordinate of the vertex is  {{{x=2}}} 



To find the y coordinate, simply plug in {{{x=2}}} into  {{{f(x)=x^2-4x-5}}} to evaluate {{{f(2)}}}




Let's evaluate {{{f(2)}}}



{{{f(x)=x^2-4x-5}}} Start with the given function.



{{{f(2)=(2)^2-4(2)-5}}} Plug in {{{x=2}}}. In other words, replace each x with 2.



{{{f(2)=(4)-4(2)-5}}} Evaluate {{{(2)^2}}} to get 4.



{{{f(2)=(4)-8-5}}} Multiply -4 and 2 to get  -8



{{{f(2)=-9}}} Now combine like terms


So the vertex is (2,-9)




Now let's graph these three points



{{{ drawing(500,500,-10,10,-10,10,
grid(1),

graph(500,500,-10,10,-10,10,0),
circle(2,-9,0.12),
circle(2,-9,0.15),


circle(5,0,0.12),
circle(5,0,0.15),


circle(-1,0,0.12),
circle(-1,0,0.15)

) }}}




Now connect the points with a parabola (note: the more points you plot the more accurate the graph is)



{{{ drawing(500,500,-10,10,-10,10,
grid(1),

graph(500,500,-10,10,-10,10,0,x^2-4x-5),
circle(2,-9,0.12),
circle(2,-9,0.15),


circle(5,0,0.12),
circle(5,0,0.15),


circle(-1,0,0.12),
circle(-1,0,0.15)

) }}}