The authors of both passages agree that King’s "I Have
1 answer:
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Change the function into vertex form:
f(x)=x²-10x+16
make a square by (adding half of -10)², which is 25, then subtract it:
f(x)=x²-10x+25-25+16
f(x)=(x-5)²-9
so the vertex is at (5,-9)
another way to do it:
f(x)=x²-10x+16
the vertex is when x=-
, b=-10, a=1 in this case.
x=-(-10/2*1)=5
plug x=5 into the quadratic function, you get f(x)=-9
f(x)=x²-10x+16
make a square by (adding half of -10)², which is 25, then subtract it:
f(x)=x²-10x+25-25+16
f(x)=(x-5)²-9
so the vertex is at (5,-9)
another way to do it:
f(x)=x²-10x+16
the vertex is when x=-
x=-(-10/2*1)=5
plug x=5 into the quadratic function, you get f(x)=-9
