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Question 9
Enter the function 
f(x)=x^(2)-18 x+6 in the form of 
f(x)=a(x-h)^(2)+k, where 
a,h, and 
k are constants. Enter your answer in the first response box.
Enter the 
x-coordinate of the vertex of the graph of 
f in the second response box.

f(x)=

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x=

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Question 99\newlineEnter the function f(x)=x218x+6 f(x)=x^{2}-18 x+6 in the form of f(x)=a(xh)2+k f(x)=a(x-h)^{2}+k , where a,h a, h , and k k are constants. Enter your answer in the first response box.\newlineEnter the x x -coordinate of the vertex of the graph of f f in the second response box.\newlinef(x)= f(x)= \newline \square \newlinex= x= \newline \square

Full solution

Q. Question 99\newlineEnter the function f(x)=x218x+6 f(x)=x^{2}-18 x+6 in the form of f(x)=a(xh)2+k f(x)=a(x-h)^{2}+k , where a,h a, h , and k k are constants. Enter your answer in the first response box.\newlineEnter the x x -coordinate of the vertex of the graph of f f in the second response box.\newlinef(x)= f(x)= \newline \square \newlinex= x= \newline \square
  1. Calculate k value: Now, calculate the value of kk by substituting hh back into the original equation.f(h)=(9)218(9)+6f(h) = (9)^2 - 18(9) + 6f(9)=81162+6f(9) = 81 - 162 + 6f(9)=75f(9) = -75So, k=75k = -75
  2. Write function in vertex form: Write the function in vertex form using the values of hh and kk.f(x)=a(xh)2+kf(x) = a(x - h)^2 + kf(x)=1(x9)275f(x) = 1(x - 9)^2 - 75
  3. Enter x-coordinate of vertex: Enter the x-coordinate of the vertex, which is the value of hh.x=9x = 9

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