Express $1+x^2-5x$ in the form of $(x+a)^2+b$. Answer ... Write down the equation of line of symmetry of the graph of $y=1+x^2-5x$. Answer ... Write down the coordinates of intersection between the line of symmetry and the graph of $y=1+x^2-5x$. Answer (..., ...)

1. Complete the square for the quadratic expression.

$1+x^2-5x=x^2-5x+1$

Half of $-5$ is $-\dfrac{5}{2}$, so

$x^2-5x+1=\left(x-\dfrac{5}{2}\right)^2-\left(\dfrac{5}{2}\right)^2+1$

$=\left(x-\dfrac{5}{2}\right)^2-\dfrac{25}{4}+1$

$=\left(x-\dfrac{5}{2}\right)^2-\dfrac{21}{4}$

1. Identify the line of symmetry from the completed square form.

From part (a),
$y=\left(x-\dfrac{5}{2}\right)^2-\dfrac{21}{4}.$
The graph is symmetric about the vertical line through its turning point, so
$x=\dfrac{5}{2}.$

1. Use the line of symmetry to get the $x$-coordinate, then find the corresponding $y$-coordinate on the graph.

From part (b), the line of symmetry is $x=\dfrac{5}{2}$.

Substitute $x=\dfrac{5}{2}$ into $y=1+x^2-5x$:

$y=1+\left(\dfrac{5}{2}\right)^2-5\left(\dfrac{5}{2}\right)$

$y=1+\dfrac{25}{4}-\dfrac{25}{2}$

$y=\dfrac{4}{4}+\dfrac{25}{4}-\dfrac{50}{4}=-\dfrac{21}{4}$

So the coordinates are $\left(\dfrac{5}{2},-\dfrac{21}{4}\right)$.