Question 130392


{{{y=1 x^2-8 x+29}}} Start with the given equation



{{{y-29=1 x^2-8 x}}}  Subtract {{{29}}} from both sides



{{{y-29=1(x^2-8x)}}} Factor out the leading coefficient {{{1}}}



Take half of the x coefficient {{{-8}}} to get {{{-4}}} (ie {{{(1/2)(-8)=-4}}}).


Now square {{{-4}}} to get {{{16}}} (ie {{{(-4)^2=(-4)(-4)=16}}})





{{{y-29=1(x^2-8x+16-16)}}} Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of {{{16}}} does not change the equation




{{{y-29=1((x-4)^2-16)}}} Now factor {{{x^2-8x+16}}} to get {{{(x-4)^2}}}



{{{y-29=1(x-4)^2-1(16)}}} Distribute



{{{y-29=1(x-4)^2-16}}} Multiply



{{{y=1(x-4)^2-16+29}}} Now add {{{29}}} to both sides to isolate y



{{{y=1(x-4)^2+13}}} Combine like terms




Now the quadratic is in vertex form {{{y=a(x-h)^2+k}}} where {{{a=1}}}, {{{h=4}}}, and {{{k=13}}}. 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 ({{{4}}},{{{13}}}) and the parabola opens upward  since {{{a>0}}}



Check:


Notice if we graph the original equation {{{y=1x^2-8x+29}}} we get:


{{{graph(500,500,-20,20,-20,20,1x^2-8x+29)}}} Graph of {{{y=1x^2-8x+29}}}. Notice how the vertex is ({{{4}}},{{{13}}}).



Notice if we graph the final equation {{{y=1(x-4)^2+13}}} we get:


{{{graph(500,500,-20,20,-20,20,1(x-4)^2+13)}}} Graph of {{{y=1(x-4)^2+13}}}. Notice how the vertex is also ({{{4}}},{{{13}}}).



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