anonymous
  • anonymous
what is the vertex of f(x)=x^2+2 what is the vertex of f(x)=x^2+2 if possible, please show steps, thank you.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
There are a number of ways: A) Plot the graph of the equation on the calculator and find the coordinates at the trough of the graph. |dw:1332891720924:dw| You can use calculus by differentiating the equation and setting the derivative to 0, then solving for x. Once the value of x is found plug that value in the original equation to get the y value: i.e. \[y=x ^{2} +2\] y' = 2x Set y' =0 => y'=2x=0 => x=0 f(0) =2 Therefore the vertex is at (0, 2) C) Use the b/2a. This gives the x value of the vertex. Plug the x value in the original equation to get the y value. i.e for an equation: \[y= ax ^{2} + bx +c \]set y =0 \[ax ^{2} + bx + C =0\] The x value of the vertex is b/2a Your equation: \[y=x ^{2} +2\] => a=1, b=0, c=2 i.e. it is equivalent to \[y=x ^{2} +0x + 2\] X value of the vertex = 0/2 = 0 substitute the x value back into the original equation to get y: thus: \[y=0^{2} + 2 =2\] The vertex is therefore (0,2).

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