Question 127713


{{{y=6 x^2+1 x+1}}} Start with the given equation



{{{y-1=6 x^2+1 x}}}  Subtract {{{1}}} from both sides



{{{y-1=6(x^2+(1/6)x)}}} Factor out the leading coefficient {{{6}}}



Take half of the x coefficient {{{1/6}}} to get {{{1/12}}} (ie {{{(1/2)(1/6)=1/12}}}).


Now square {{{1/12}}} to get {{{1/144}}} (ie {{{(1/12)^2=(1/12)(1/12)=1/144}}})





{{{y-1=6(x^2+(1/6)x+1/144-1/144)}}} Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of {{{1/144}}} does not change the equation




{{{y-1=6((x+1/12)^2-1/144)}}} Now factor {{{x^2+(1/6)x+1/144}}} to get {{{(x+1/12)^2}}}



{{{y-1=6(x+1/12)^2-6(1/144)}}} Distribute



{{{y-1=6(x+1/12)^2-1/24}}} Multiply



{{{y=6(x+1/12)^2-1/24+1}}} Now add {{{1}}} to both sides to isolate y



{{{y=6(x+1/12)^2+23/24}}} Combine like terms




Now the quadratic is in vertex form {{{y=a(x-h)^2+k}}} where {{{a=6}}}, {{{h=-1/12}}}, and {{{k=23/24}}}. Remember (h,k) is the vertex and "a" is the stretch/compression factor. Also "a" tells us which direction the parabola opens.




So in this case the vertex is ({{{-1/12}}},{{{23/24}}}) and the parabola opens upward  since {{{a>0}}}



Check:


Notice if we graph the original equation {{{y=6x^2+1x+1}}} we get:


{{{graph(500,500,-10,10,-10,10,6x^2+1x+1)}}} Graph of {{{y=6x^2+1x+1}}}. Notice how the vertex is ({{{-1/12}}},{{{23/24}}}).



Notice if we graph the final equation {{{y=6(x+1/12)^2+23/24}}} we get:


{{{graph(500,500,-10,10,-10,10,6(x+1/12)^2+23/24)}}} Graph of {{{y=6(x+1/12)^2+23/24}}}. Notice how the vertex is also ({{{-1/12}}},{{{23/24}}}).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.