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Algebra 12 Online
OpenStudy (the_rebel_outlaw):

When written in standard form, the quadratic equation, x2 - 2x - 35 = 0, is equivalent to y = (x - 1)2 - 36. What is the vertex of the graph of the quadratic equation, x2 - 2x - 35 = 0?

OpenStudy (nincompoop):

:)

Nnesha (nnesha):

if the equation is written in standard form \[y=a(x-h)^2+k\] (h,k) is the vertex point \[\large\rm y=(x-1)^2-36\]

OpenStudy (the_rebel_outlaw):

can i get an example please

OpenStudy (nincompoop):

if you are not given the vertex form, you can always calculate it by using \(h=\dfrac{-b}{2a} \) then k would be solved by \(k = \dfrac{4ac-b^2}{4a} \)

Nnesha (nnesha):

\[\large\rm y=a(x-\color{Red}{h})^2 + \color{blue}{k}\] for example \[\large\rm y=a(x+2 )^2 +k\] as you can see if we rewrite in standard form it should be `x-h` \[\large\rm y=a(x-\color{Red}{(-2)} )^2 +\color{blue}{3}\] \[(\color{ReD}{h},\color{blue}{k})\] \[(\color{ReD}{-2},\color{blue}{3})\] (-2,3) is the vertex

Nnesha (nnesha):

\[\large\rm Ax^\color{Red}{2}+Bx+C=0\]

OpenStudy (the_rebel_outlaw):

oh ok thank you

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