please help ill give medal y=x^2-2x+1 A. 1 B. –1 C. –2 D. 2
do you wanna x of vertex ( axis of symmetry)?
Use the zeros of the following quadratic to find the x-value of the vertex.
thats simple put all the stuff in quadratic formula i.e a= 1 , b= -2, c=1 in (-b +- underroot (b^2 - 4 ac )) / 2a root will be 1
you lost me at this part (-b +- underroot (b^2 - 4 ac )) / 2a
I forgot latex just hold on try to show you equation
try to turn it in standard form y=a(x-b)^2+b
y=x^2-2x+1=(x-)^2 so x of vertex is x=1
put x=1 in it to find y of vertex
\[\frac{ -b \pm \sqrt{b^{2}- 4ac} }{ 2a }\]
thats quadratic formula :)
annas you say these are the roots of quadratic
lol smart guy y= (1)^2 -2 (1) + 1 = 1 - 2 +1= 2 - 2 = 0 xD not an option
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