Question 127805
a)




{{{y=10 x^2+1 x-1}}} Start with the given equation



{{{y+1=10 x^2+1 x}}} Add {{{1}}} to both sides



{{{y+1=10(x^2+(1/10)x)}}} Factor out the leading coefficient {{{10}}}



Take half of the x coefficient {{{1/10}}} to get {{{1/20}}} (ie {{{(1/2)(1/10)=1/20}}}).


Now square {{{1/20}}} to get {{{1/400}}} (ie {{{(1/20)^2=(1/20)(1/20)=1/400}}})





{{{y+1=10(x^2+(1/10)x+1/400-1/400)}}} Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of {{{1/400}}} does not change the equation




{{{y+1=10((x+1/20)^2-1/400)}}} Now factor {{{x^2+(1/10)x+1/400}}} to get {{{(x+1/20)^2}}}



{{{y+1=10(x+1/20)^2-10(1/400)}}} Distribute



{{{y+1=10(x+1/20)^2-1/40}}} Multiply



{{{y=10(x+1/20)^2-1/40-1}}} Now add {{{+1}}} to both sides to isolate y



{{{y=10(x+1/20)^2-41/40}}} Combine like terms




Now the quadratic is in vertex form {{{y=a(x-h)^2+k}}} where {{{a=10}}}, {{{h=-1/20}}}, and {{{k=-41/40}}}. 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 ({{{-1/20}}},{{{-41/40}}}) and the parabola opens upward  since {{{a>0}}}



Check:


Notice if we graph the original equation {{{y=10x^2+1x-1}}} we get:


{{{graph(500,500,-10,10,-10,10,10x^2+1x-1)}}} Graph of {{{y=10x^2+1x-1}}}. Notice how the vertex is ({{{-1/20}}},{{{-41/40}}}).



Notice if we graph the final equation {{{y=10(x+1/20)^2-41/40}}} we get:


{{{graph(500,500,-10,10,-10,10,10(x+1/20)^2-41/40)}}} Graph of {{{y=10(x+1/20)^2-41/40}}}. Notice how the vertex is also ({{{-1/20}}},{{{-41/40}}}).



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




c)


The min or max occurs at the vertex and the max/min value is the y value of the vertex. So in this case the minimum is {{{-41/40}}}