Chapter 10.5, Problem 17E

To determine

Find the vertex coordinates, focus, the equations of the directrix and axis of symmetry of the given parabola and draw a graph.

Answer to Problem 17E

Vertex $\to \left(h,k\right)=\left(2,-2\right)$

Focus $\to \left(h,k+p\right)=\left(2,-2+\frac{1}{4}\right)=\left(2,-\frac{7}{4}\right)$

Directrix $\to$

  $y=-\frac{9}{4}$

Axis of symmetry $\to$

  $x=2$

Explanation of Solution

Given:

The given equation of the parabola is $y-2=x^2-4x$ .

Calculation:

The standard equation for the up-down facing parabola is

  $4p\left(y-k\right)={\left(x-h\right)}^2$

Where $\left(h,k\right)$ is the vertex and $\left|p\right|$ is the focal length.

Write the given equation in standard form.

  $\begin{array}{l}y-2=x^2-4x\\ \Rightarrow y-2+4=x^2-4x+4\\ \Rightarrow y+2={\left(x-2\right)}^2\\ \Rightarrow 4\cdot \frac{1}{4}\left[y-\left(-2\right)\right]={\left(x-2\right)}^2\end{array}$

Vertex $\to \left(h,k\right)=\left(2,-2\right)$

Parabola is symmetric around the y-axis.

Focus $\to \left(h,k+p\right)=\left(2,-2+\frac{1}{4}\right)=\left(2,-\frac{7}{4}\right)$

The directrix is a line parallel to the x-axis, because parabola is symmetric around the y-axis.

  $\begin{array}{l}y=-2-p\\ \Rightarrow y=-2-\frac{1}{4}\\ \Rightarrow y=-\frac{9}{4}\end{array}$

The axis of symmetry is a line parallel to the y-axis which intersects the vertex.

  $x=2$

Draw a table for the value of $y$ with respect to $x$ .

$x$	$y=x^2-4x+2$	$\left(x,y\right)$
$\begin{array}{l}-1\\ 0\\ 2\\ 4\\ 5\end{array}$	$\begin{array}{l}7\\ 2\\ -2\\ 2\\ 7\end{array}$	$\begin{array}{l}\left(-1,7\right)\\ \left(0,2\right)\\ \left(2,-2\right)\\ \left(4,2\right)\\ \left(5,7\right)\end{array}$

Hence the graph of the parabola is given below.