Chapter EP, Problem 5.1.3EP

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 5.1.3EP

y -intercept is $5$.

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 $f\left(x\right)=x^2+5$.

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.

  $f\left(0\right)={\left(0\right)}^2+5=5$

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 5.1.3EP

$x$	$f\left(x\right)$	$\left(x,f\left(x\right)\right)$
$0$	$5$	$\left(0,5\right)$
$1$	$6$	$\left(1,6\right)$
$-1$	$6$	$\left(-1,6\right)$
$2$	$9$	$\left(2,9\right)$
$-2$	$9$	$\left(-2,9\right)$
$3$	$14$	$\left(3,14\right)$
$-3$	$14$	$\left(-3,14\right)$

Explanation of Solution

Given information:

The function provided is $f\left(x\right)=x^2+5$.

Calculation:

Plot some points on the given quadratic function as shown:

  $\begin{array}{l}x=0,f\left(0\right)={\left(0\right)}^2+5=5\\ x=-1,f\left(-1\right)={\left(-1\right)}^2+5=6\\ x=1,f\left(1\right)={\left(1\right)}^2+5=6\\ x=-2,f\left(-2\right)={\left(-2\right)}^2+5=9\\ x=2,f\left(2\right)={\left(2\right)}^2+5=9\\ x=-3,f\left(-3\right)={\left(-3\right)}^2+5=14\\ x=3,f\left(3\right)={\left(3\right)}^2+5=14\end{array}$

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

$x$	$f\left(x\right)$	$\left(x,f\left(x\right)\right)$
$0$	$5$	$\left(0,5\right)$
$1$	$6$	$\left(1,6\right)$
$-1$	$6$	$\left(-1,6\right)$
$2$	$9$	$\left(2,9\right)$
$-2$	$9$	$\left(-2,9\right)$
$3$	$14$	$\left(3,14\right)$
$-3$	$14$	$\left(-3,14\right)$

c.

To determine

To graph: the given quadratic function.

Explanation of Solution

Given information:

The function provided is $f\left(x\right)=x^2+5$.

Calculation:

Some points on the quadratic function are as shown below:

$x$	$f\left(x\right)$	$\left(x,f\left(x\right)\right)$
$0$	$5$	$\left(0,5\right)$
$1$	$6$	$\left(1,6\right)$
$-1$	$6$	$\left(-1,6\right)$
$2$	$9$	$\left(2,9\right)$
$-2$	$9$	$\left(-2,9\right)$
$3$	$14$	$\left(3,14\right)$
$-3$	$14$	$\left(-3,14\right)$

Construct the graph of the function by plotting the points.