A quadratic function f(x)=x2−2x+2.
a.) Find the quadratic function in standard form.
f(x)=x2−2x+2f(x)=1(x2−2x)+2Factor out 1 from x-termsf(x)=1(x2−2x+1)+2−(1)(1)Complete the square: add 1 inside parentheses, subtract (1)(1) outsidef(x)=(x−1)2+1Factor and simplify
The standard form is f(x)=(x−1)2+1.
b.) Find its vertex and its x and y-intercepts.
By using f(x)=a(x−h)2+k with vertex at (h,k).
The vertex of the function f(x)=(x−1)2+1 is at (1,1).
Solving for x-interceptSolving for y-interceptWe set f(x)=0, thenWe set x=0, then0=(x−1)2+1Subtract 1y=(0−1)2+1Substitute x=0−1=(x−1)2Take the square rooty=1+1Simplify±√−1=x−1y=2x-intercept does not exist
c.) Draw its graph.
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