<|system|>
Find the distance between the vertex of the graph of the equation $f(x) = x^2 - 8x + 15$ and the point $(0, 2)$.</s>.
<|user|>
Can you answer based on context?
<|assistant|>
I know that the vertex of a quadratic function is the point where the function has its maximum or minimum value, depending on the sign of the coefficient of $x^2$. In this case, the coefficient of $x^2$ is positive, so the function has a minimum value at the vertex. To find the vertex, I can use the formula $x = -\frac{b}{2a}$, where $a$ and $b$ are the coefficients of $x^2$ and $x$ respectively. In this case, $a = 1$ and $b = -8$, so the $x$-coordinate of the vertex is $x = -\frac{-8}{2 \cdot 1} = 4$. To find the $y$-coordinate of the vertex, I can plug in $x = 4$ into the function: $f(4) = 4^2 - 8(4) + 15 = 16 - 32 + 15 = -1$. So the vertex is $(4, -1)$. Now I need to find the distance between this point and $(0, 2)$. To do this, I can use the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, where $(x_1, y_1)$ and $(x_2, y_2)$ are the two points. In this case, $(x_1, y_1) = (0, 2)$ and $(x_2, y_2) = (4, -1)$, so I plug these values into the formula: $d = \sqrt{(4 - 0)^2 + (-1 - 2)^2}$. I simplify the expression inside the square root: $d = \sqrt{16 + 9} = \sqrt{25} = 5$. Therefore, the distance between the vertex and the point is 5 units.

