Question 246034
Part 1) Find the vertices of {{{f(x)=-x^2+3x+10}}} and {{{g(x)=2x^2+2x+11/4}}}


part a) Let's find the vertex of {{{f(x)=-x^2+3x+10}}}




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+3x+10}}}, we can see that {{{a=-1}}}, {{{b=3}}}, and {{{c=10}}}.



{{{x=(-(3))/(2(-1))}}} Plug in {{{a=-1}}} and {{{b=3}}}.



{{{x=(-3)/(-2)}}} Multiply 2 and {{{-1}}} to get {{{-2}}}.



{{{x=3/2}}} Reduce.



So the x-coordinate of the vertex is {{{x=3/2}}}. Note: this means that the axis of symmetry is also {{{x=3/2}}}.



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+3x+10}}} Start with the given equation.



{{{y=-(3/2)^2+3(3/2)+10}}} Plug in {{{x=3/2}}}.



{{{y=-1(9/4)+3(3/2)+10}}} Square {{{3/2}}} to get {{{9/4}}}.



{{{y=-9/4+3(3/2)+10}}} Multiply {{{-1}}} and {{{9/4}}} to get {{{-9/4}}}.



{{{y=-9/4+9/2+10}}} Multiply {{{3}}} and {{{3/2}}} to get {{{9/2}}}.



{{{y=49/4}}} Combine like terms.



So the y-coordinate of the vertex is {{{y=49/4}}}.



So the vertex is *[Tex \LARGE \left(\frac{3}{2},\frac{49}{4}\right)].



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b) Now let's find the vertex of {{{g(x)=2x^2+2x+11/4}}}





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=2x^2+2x+11/4}}}, we can see that {{{a=2}}}, {{{b=2}}}, and {{{c=11/4}}}.



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



{{{x=(-2)/(4)}}} Multiply 2 and {{{2}}} to get {{{4}}}.



{{{x=-1/2}}} Reduce.



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=2x^2+2x+11/4}}} Start with the given equation.



{{{y=2(-1/2)^2+2(-1/2)+11/4}}} Plug in {{{x=-1/2}}}.



{{{y=2(1/4)+2(-1/2)+11/4}}} Square {{{-1/2}}} to get {{{1/4}}}.



{{{y=1/2+2(-1/2)+11/4}}} Multiply {{{2}}} and {{{1/4}}} to get {{{1/2}}}.



{{{y=1/2-1+11/4}}} Multiply {{{2}}} and {{{-1/2}}} to get {{{-1}}}.



{{{y=9/4}}} Combine like terms.



So the y-coordinate of the vertex is {{{y=9/4}}}.



So the vertex is *[Tex \LARGE \left(-\frac{1}{2},\frac{9}{4}\right)].



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So to recap, the vertices of {{{f(x)=-x^2+3x+10}}} and {{{g(x)=2x^2+2x+11/4}}} are *[Tex \LARGE \left(\frac{3}{2},\frac{49}{4}\right)] and *[Tex \LARGE \left(-\frac{1}{2},\frac{9}{4}\right)] respectively.



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Part 2) Now use the distance formula to find the distance between the two vertices (which are essentially points)




Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(\frac{3}{2},\frac{49}{4}\right)]. So this means that {{{x[1]=3/2}}} and {{{y[1]=49/4}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(-\frac{1}{2},\frac{9}{4}\right)].  So this means that {{{x[2]=-1/2}}} and {{{y[2]=9/4}}}.




{{{d=sqrt((x[1]-x[2])^2+(y[1]-y[2])^2)}}} Start with the distance formula.



{{{d=sqrt((3/2--1/2)^2+(49/4-9/4)^2)}}} Plug in {{{x[1]=3/2}}},  {{{x[2]=-1/2}}}, {{{y[1]=49/4}}}, and {{{y[2]=9/4}}}.



{{{d=sqrt((2)^2+(49/4-9/4)^2)}}} Subtract {{{-1/2}}} from {{{3/2}}} to get {{{3/2--1/2=4/2=2}}}.



{{{d=sqrt((2)^2+(10)^2)}}} Subtract {{{9/4}}} from {{{49/4}}} to get {{{49/4-9/4=40/4=10}}}.



{{{d=sqrt(4+(10)^2)}}} Square {{{2}}} to get {{{4}}}.



{{{d=sqrt(4+100)}}} Square {{{10}}} to get {{{100}}}.



{{{d=sqrt(104)}}} Add {{{4}}} to {{{100}}} to get {{{104}}}.



{{{d=2*sqrt(26)}}} Simplify the square root.



So our answer is {{{d=2*sqrt(26)}}} 



So the exact distance between the two vertices is {{{2*sqrt(26)}}} units.