+37097 vertex form of a parabola's equation is generally expressed as: y = a(x-h)2+k
Where the vertex is h, k so for this one the vertex is (3,1)
+14986 ElectricPavlov, please help me.
How do you get to (3,1)?
Greeting asinus :- )
+37097 y=2(x-3)²+1 This is the vertex form of a parabola expressed as: y = a(x-h)2+k (Where the vertex is h, k)
y=a(x-h)^2 +k so -h = -3 or h=3 and k= 1 so the vertex is (3,1)
+14986 which point is the vertex of y=2(x-3)²+1
\(y=f(x)=2(x-3)^2+1\)
\(f(x)=2(x^2-6x+9)+1 =2x^2-12x+19\)
1. Derivation = 0
\(f'(x)=4x-12=0 \) Change of page with change of sign
\(x_{min}=3\) set x
\(y=2x^2-12x+19=18-36+19=1\)
\(y_{min}=1\)
The vertex of the function is P (3;1)
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