Question 118068
o 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+x-6}}} we can see that a=1 and b=1


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



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



So the axis of symmetry is  {{{x=-1/2}}}



So the x-coordinate of the vertex is {{{x=-1/2}}} (which is {{{x=-0.5}}} in decimal form). Lets plug this into the equation to find the y-coordinate of the vertex.



Lets evaluate {{{g(-0.5)}}}


{{{g(x)=x^2+x-6}}} Start with the given polynomial



{{{g(-0.5)=(-0.5)^2+(-0.5)-6}}} Plug in {{{x=-0.5}}}



{{{g(-0.5)=(0.25)+(-0.5)-6}}} Raise -0.5 to the second power to get 0.25



{{{g(-0.5)=-6.25}}} Combine like terms



So the vertex is (-0.5,-6.25)




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To find the x-intercepts, set the whole function equal to zero and solve for x.




{{{x^2+x-6=0}}} Start with the given equation


{{{(x+3)(x-2)=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+3=0}}} or  {{{x-2=0}}} 


{{{x=-3}}} or  {{{x=2}}}    Now solve for x in each case



So our answer is 

 {{{x=-3}}} or  {{{x=2}}} 



Notice if we graph {{{y=x^2+x-6}}} we can see that the roots are {{{x=-3}}} and  {{{x=2}}}. We can also see that the vertex is (-0.5,-6.25). So this visually verifies our answer.



{{{drawing(500,500,-10,10,-10,10, 
graph(500,500,-10,10,-10,10,0, x^2+x-6),
circle(-0.5,-6.25,0.08),
circle(-0.5,-6.25,0.1),
circle(-0.5,-6.25,0.12)
) }}} Graph of {{{y=x^2+x-6}}} with the x-intercepts {{{x=-3}}} and  {{{x=2}}} and the vertex (-0.5,-6.25).