Chapter SH, Problem 4.1EP

a.

To determine

Find the y-intercept, axis of symmetry and the x-coordinate of the vertex.

Answer to Problem 4.1EP

The y-intercept $=-3$

Theequation of the axis of symmetry is $x=0$

The x-coordinate of the vertex $=0$

Explanation of Solution

Given:

The given function is $f\left(x\right)=x^2-3$ .

Calculation:

  $f\left(x\right)=x^2-3$

The function is of the form $f\left(x\right)=a{x}^2+bx+c$ .

  $a=1,b=0,c=-3$

y-intercept $=c=-3$ .

Equation of the axis of symmetry.

  $\begin{array}{l}x=-\frac{b}{2a}\\ \Rightarrow x=-\frac{0}{2\cdot 1}\\ \Rightarrow x=0\end{array}$

The axis of symmetry intersects a parabola at only one point, called the vertex.

The x-coordinate of the vertex $=-\frac{b}{2a}$

  $=-\frac{0}{2\cdot 1}$

  $=0$

Hence

the y-intercept $=-3$

theequation of the axis of symmetry is $x=0$

the x-coordinate of the vertex $=0$

b.

To determine

Create a table for the values with vertex.

Answer to Problem 4.1EP

The table for the values of $y$ with respect to $x$ .

$x$	$y=x^2-3$	$\left(x,y\right)$
$\begin{array}{l}-3\\ -2\\ 0\\ 1\\ 2\\ 3\end{array}$	$\begin{array}{l}6\\ 1\\ -3\\ -2\\ 1\\ 6\end{array}$	$\begin{array}{l}\left(-3,6\right)\\ \left(-2,1\right)\\ \left(0,-3\right)\\ \left(1,-2\right)\\ \left(2,1\right)\\ \left(3,6\right)\end{array}$

Explanation of Solution

Given:

The given function is $f\left(x\right)=x^2-3$ .

Calculation:

  $f\left(x\right)=x^2-3$

The x-coordinate of the vertex $=-\frac{b}{2a}$

  $=-\frac{0}{2\cdot 1}$

  $=0$

Substitute the value of $x=0$ in the given function.

  $y=0^2-3=-3$

Vertex $=\left(0,-3\right)$

The table for the values of $y$ with respect to $x$ is given below.

$x$	$y=x^2-3$	$\left(x,y\right)$
$\begin{array}{l}-3\\ -2\\ 0\\ 1\\ 2\\ 3\end{array}$	$\begin{array}{l}6\\ 1\\ -3\\ -2\\ 1\\ 6\end{array}$	$\begin{array}{l}\left(-3,6\right)\\ \left(-2,1\right)\\ \left(0,-3\right)\\ \left(1,-2\right)\\ \left(2,1\right)\\ \left(3,6\right)\end{array}$

c.

To determine

Find the y-intercept, axis of symmetry and the x-coordinate of the vertex.

Explanation of Solution

Given:

The given function is $f\left(x\right)=x^2-3$ .

Calculation:

  $f\left(x\right)=x^2-3$

The function is of the form $f\left(x\right)=a{x}^2+bx+c$ .

  $a=1,b=0,c=-3$

y-intercept $=c=-3$ .

Equation of the axis of symmetry.

  $\begin{array}{l}x=-\frac{b}{2a}\\ \Rightarrow x=-\frac{0}{2\cdot 1}\\ \Rightarrow x=0\end{array}$

The axis of symmetry intersects a parabola at only one point, called the vertex.

The x-coordinate of the vertex $=-\frac{b}{2a}$

  $=-\frac{0}{2\cdot 1}$

  $=0$

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

$x$	$y=x^2-3$	$\left(x,y\right)$
$\begin{array}{l}-3\\ -2\\ 0\\ 1\\ 2\\ 3\end{array}$	$\begin{array}{l}6\\ 1\\ -3\\ -2\\ 1\\ 6\end{array}$	$\begin{array}{l}\left(-3,6\right)\\ \left(-2,1\right)\\ \left(0,-3\right)\\ \left(1,-2\right)\\ \left(2,1\right)\\ \left(3,6\right)\end{array}$

Hence the graph of the function is given below.