For a given problem $g(x) = 5x^2 - 2x +1$, we were expected to describe the graph. WolframAlpha and I are finding conflicting issues. I do not think I made an approximation anywhere that would explain the difference.
For a given problem $g(x) = 5x^2 -2x +1$, the problem is already in the form of $y = ax^2 +bx + c$ If we can identify the vertex coordinates $(h, k)$ using $h = -b/2a$ and $k = f (-b/2a)$ If $h = -b/2a$ for given problem is $-(-2)/2(5) = 4/10 = 2/5 = h$ If $k = f (-b/2a)$ (or in our case, $g(-b/2a) = g (2/5) = 5(2/5)^2 - 2(2/5) +1 = 5 (4/25) - (4/5) +1 = 20/25 - 4/5 + 1 = 4/5 - 4/5 +1 = 1$ Then vertex coordinates $(h, k) = (2/5, 1)$
But if I enter g(x) = 5x^2 -2x +1 in WolframAlpha, it says (h, k) is (1/5, 4/5). What am I missing in my work that is leading to this problem over and over again? There is a simple error I cannot find which is very embarrassing.
You have an error, $$h=\frac{-b}{2a}=\frac{-(-2)}{2*5}=\frac{2}{10}=\frac{1}{5}$$
You make:
$$h=\frac{-2b}{2a}=\frac{-(-2*2)}{10}=\frac{4}{10}=\frac{2}{5}$$
So, the true value of $h$ is $h=\frac{1}{5}$ and $f(h)=\frac{4}{5}$ that is the answer Wolfram Alpha give you.