Chapter EP, Problem 5.1.4EP

a.

To determine

To find:the y -intercept, the equation of the axis of symmetry, and the x -coordinate of the vertex for the given quadratic function.

Answer to Problem 5.1.4EP

y -intercept is $-2$.

The equation of the axis of symmetry is $x=0$.

x -coordinate of the vertex is $0$.

Explanation of Solution

Given information:

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

Formula used:

For the quadratic function $f\left(x\right)=a{x}^2+bx+c$

Formula to compute the x -coordinate of the vertex is $x=-\frac{b}{2\times a}$ .

‘a’ and ‘b’ are coefficients $x^2$ and $x$ respectively.

For y -intercept,

Substitute $x=0$ in the function.

Formula for axis of symmetry.

  $x=-\frac{b}{2\times a}$

Calculation :

For y -intercept,

Substitute $x=0$ in the function.

  $f\left(0\right)=-{\left(0\right)}^2-2=-2$

x -coordinate of the vertex is

  $x=-\frac{b}{2\times a}$

Here, $a=-1$ , $b=0$ and $c=5$.

  $\begin{array}{c}x=-\frac{0}{2\times \left(-1\right)}\\ =0\end{array}$

Formula for axis of symmetry.

  $\begin{array}{c}x=-\frac{0}{2\times \left(-1\right)}\\ =0\end{array}$

b.

To determine

To plot: some points on the given quadratic function.

Answer to Problem 5.1.4EP

$x$	$f\left(x\right)$	$\left(x,f\left(x\right)\right)$
$0$	$-2$	$\left(0,-2\right)$
$1$	$-3$	$\left(1,-3\right)$
$-1$	$-3$	$\left(-1,-3\right)$
$2$	$-6$	$\left(2,-6\right)$
$-2$	$-6$	$\left(-2,-6\right)$
$3$	$-11$	$\left(3,-11\right)$
$-3$	$-11$	$\left(-3,-11\right)$

Explanation of Solution

Given information:

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

Calculation:

Plot some points on the given quadratic function as shown:

  $\begin{array}{l}x=0,f\left(0\right)=-{\left(0\right)}^2-2=-2\\ x=-1,f\left(-1\right)=-{\left(-1\right)}^2-2=-3\\ x=1,f\left(1\right)=-{\left(1\right)}^2-2=-3\\ x=-2,f\left(-2\right)=-{\left(-2\right)}^2-2=-6\\ x=2,f\left(2\right)=-{\left(2\right)}^2-2=-6\\ x=-3,f\left(-3\right)=-{\left(-3\right)}^2-2=-11\\ x=3,f\left(3\right)=-{\left(3\right)}^2-2=-11\end{array}$

So, the points can be constructed in a table as shown:

$x$	$f\left(x\right)$	$\left(x,f\left(x\right)\right)$
$0$	$-2$	$\left(0,-2\right)$
$1$	$-3$	$\left(1,-3\right)$
$-1$	$-3$	$\left(-1,-3\right)$
$2$	$-6$	$\left(2,-6\right)$
$-2$	$-6$	$\left(-2,-6\right)$
$3$	$-11$	$\left(3,-11\right)$
$-3$	$-11$	$\left(-3,-11\right)$

c.

To determine

To graph: the given quadratic function.

Explanation of Solution

Given information:

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

Calculation:

Some points on the quadratic function are as shown below:

$x$	$f\left(x\right)$	$\left(x,f\left(x\right)\right)$
$0$	$-2$	$\left(0,-2\right)$
$1$	$-3$	$\left(1,-3\right)$
$-1$	$-3$	$\left(-1,-3\right)$
$2$	$-6$	$\left(2,-6\right)$
$-2$	$-6$	$\left(-2,-6\right)$
$3$	$-11$	$\left(3,-11\right)$
$-3$	$-11$	$\left(-3,-11\right)$

Construct the graph of the function by plotting the points.