Question 154223
y = 3x^2 + 12x + 9 ... Start with the given equation.



y' = 6x + 12 ... Find the first derivative



y'' = 6 ... Find the second derivative 



Since y'' is <b>always</b> positive, this means that the function is concave up.



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b)


Y-Intercept:


{{{y=3x^2 + 12x + 9}}} Start with the given equation.



{{{y=3(0)^2 + 12(0) + 9}}} Plug in {{{x=0}}}.



{{{y=3(0)+12(0)+9}}} Square {{{0}}} to get {{{0}}}.



{{{y=0+12(0)+9}}} Multiply {{{3}}} and {{{0}}} to get {{{0}}}.



{{{y=0+0+9}}} Multiply {{{12}}} and {{{0}}} to get {{{0}}}.



{{{y=9}}} Combine like terms.



So the y-intercept is (0,9)



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c)


X-Intercept(s):



{{{y=3x^2 + 12x + 9}}} Start with the given equation.



{{{0=3x^2 + 12x + 9}}} Plug in {{{y=0}}}



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=3}}}, {{{b=12}}}, and {{{c=9}}}



Let's use the quadratic formula to solve for x



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



{{{x = (-(12) +- sqrt( (12)^2-4(3)(9) ))/(2(3))}}} Plug in  {{{a=3}}}, {{{b=12}}}, and {{{c=9}}}



{{{x = (-12 +- sqrt( 144-4(3)(9) ))/(2(3))}}} Square {{{12}}} to get {{{144}}}. 



{{{x = (-12 +- sqrt( 144-108 ))/(2(3))}}} Multiply {{{4(3)(9)}}} to get {{{108}}}



{{{x = (-12 +- sqrt( 36 ))/(2(3))}}} Subtract {{{108}}} from {{{144}}} to get {{{36}}}



{{{x = (-12 +- sqrt( 36 ))/(6)}}} Multiply {{{2}}} and {{{3}}} to get {{{6}}}. 



{{{x = (-12 +- 6)/(6)}}} Take the square root of {{{36}}} to get {{{6}}}. 



{{{x = (-12 + 6)/(6)}}} or {{{x = (-12 - 6)/(6)}}} Break up the expression. 



{{{x = (-6)/(6)}}} or {{{x =  (-18)/(6)}}} Combine like terms. 



{{{x = -1}}} or {{{x = -3}}} Simplify. 



So the answers are {{{x = -1}}} or {{{x = -3}}} 


  
So the x-intercepts are (-1,0) and (-3,0)



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d)


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=3x^2+12x+9}}}, we can see that {{{a=3}}}, {{{b=12}}}, and {{{c=9}}}.



{{{x=(-(12))/(2(3))}}} Plug in {{{a=3}}} and {{{b=12}}}.



{{{x=(-12)/(6)}}} Multiply 2 and {{{3}}} to get {{{6}}}.



{{{x=-2}}} Divide.



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



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



{{{y=3x^2+12x+9}}} Start with the given equation.



{{{y=3(-2)^2+12(-2)+9}}} Plug in {{{x=-2}}}.



{{{y=3(4)+12(-2)+9}}} Square {{{-2}}} to get {{{4}}}.



{{{y=12+12(-2)+9}}} Multiply {{{3}}} and {{{4}}} to get {{{12}}}.



{{{y=12-24+9}}} Multiply {{{12}}} and {{{-2}}} to get {{{-24}}}.



{{{y=-3}}} Combine like terms.



So the y-coordinate of the vertex is {{{y=-3}}}.



So the vertex is *[Tex \LARGE \left(-2,-3\right)].



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e)


Maximum or Minimum value:



Since the function is concave up, this means that the function has a minimum. The max/min value correspond to the y coordinate of the vertex. So the minimum value is {{{y=-3}}}



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Here's a graph to verify our answers:



{{{ graph( 500, 500, -10, 10, -10, 10, 3x^2+12x+9) }}} Graph of {{{y=3x^2+12x+9}}}