Question 150542
Vertex:




In order to find the vertex, we first need to find the x-coordinate of the vertex.



To find the x-coordinate of the vertex, use this formula: {{{x=(-b)/(2a)}}}.



{{{x=(-b)/(2a)}}} Start with the given formula.



From {{{y=x^2+x-6}}}, we can see that {{{a=1}}}, {{{b=1}}}, and {{{c=-6}}}.



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



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



So the x-coordinate of the vertex is {{{x=-1/2}}}. Note: this means that the axis of symmetry is also {{{x=-1/2}}}.



Now that we know the x-coordinate of the vertex, we can use it to find the y-coordinate of the vertex.



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



{{{y=(-1/2)^2-1/2-6}}} Plug in {{{x=-1/2}}}.



{{{y=1/4-1/2-6}}} Square {{{-1/2}}} to get {{{1/4}}}.



{{{y=-25/4}}} Combine the fractions.



So the y-coordinate of the vertex is {{{y=-25/4}}}.



So the vertex is *[Tex \LARGE \left(-\frac{1}{2},-\frac{25}{4}\right)] which in decimal form is (-0.5, -6.25).




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Intercepts:


X-intercept(s):


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



{{{0=x^2+x-6}}} Plug in {{{y=0}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(1) +- sqrt( (1)^2-4(1)(-6) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=1}}}, and {{{c=-6}}}



{{{x = (-1 +- sqrt( 1-4(1)(-6) ))/(2(1))}}} Square {{{1}}} to get {{{1}}}. 



{{{x = (-1 +- sqrt( 1--24 ))/(2(1))}}} Multiply {{{4(1)(-6)}}} to get {{{-24}}}



{{{x = (-1 +- sqrt( 1+24 ))/(2(1))}}} Rewrite {{{sqrt(1--24)}}} as {{{sqrt(1+24)}}}



{{{x = (-1 +- sqrt( 25 ))/(2(1))}}} Add {{{1}}} to {{{24}}} to get {{{25}}}



{{{x = (-1 +- sqrt( 25 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (-1 +- 5)/(2)}}} Take the square root of {{{25}}} to get {{{5}}}. 



{{{x = (-1 + 5)/(2)}}} or {{{x = (-1 - 5)/(2)}}} Break up the expression. 



{{{x = (4)/(2)}}} or {{{x =  (-6)/(2)}}} Combine like terms. 



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



So the solutions are {{{x = 2}}} or {{{x = -3}}} 

  
This means that the x-intercepts are (-3,0) and (2,0)


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Y-intercept:


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



{{{y=(0)^2+0-6}}} Plug in {{{x=0}}}.



{{{y=1(0)+0-6}}} Square {{{0}}} to get {{{0}}}.



{{{y=0+0-6}}} Multiply {{{1}}} and {{{0}}} to get {{{0}}}.



{{{y=-6}}} Combine like terms.



So the y-intercept is (0,-6)



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Summary:


So the vertex is *[Tex \LARGE \left(-\frac{1}{2},-\frac{25}{4}\right)] which in decimal form is (-0.5, -6.25)


The x-intercepts are (-3,0) and (2,0) and the  y-intercept is (0,-6)




Plot these points on the coordinate system


{{{ drawing(900, 900, -10, 10, -10, 10,
grid(1),
graph(900, 900, -10, 10, -10, 10, 0),

circle(-3,0,0.08),circle(-3,0,0.10),

circle(0,-6,0.08),circle(0,-6,0.10),

circle(-1/2,-25/4,0.08),circle(-1/2,-25/4,0.10),

circle(2,0,0.08),circle(2,0,0.10)



)}}}


<a name="graph">



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<h4>Graph:</h4>

Now draw a parabola through all of the points to graph {{{y=x^2+x-6}}}:



{{{ drawing(900, 900, -10, 10, -10, 10,
grid(1),
graph(900, 900, -10, 10, -10, 10, x^2+x-6),
circle(-3,0,0.08),circle(-3,0,0.10),

circle(0,-6,0.08),circle(0,-6,0.10),

circle(-1/2,-25/4,0.08),circle(-1/2,-25/4,0.10),

circle(2,0,0.08),circle(2,0,0.10)

)}}} Graph of {{{y=x^2+x-6}}}