Chapter EP, Problem 5.1.7EP

(a)

To determine

To Find: The y intercept of the equation of the axis of symmetry and the x coordinate of the vertex.

Answer to Problem 5.1.7EP

The value of the y intercept is $-3$ and equation of the symmetry is $3$ then the x coordinate of the vertex is $3$ .

Explanation of Solution

Given:

The given equation is $f\left(x\right)=x^2+6x-3$ .

Calculation:

Consider the given equation is,

  $f\left(x\right)=x^2+6x-3$

Then, find the y intercept as,

  $\begin{array}{c}f\left(0\right)={\left(0\right)}^2+6\left(0\right)-3\\ =-3\end{array}$

The axis of the symmetry is obtained as,

  $\begin{array}{c}x=-\frac{b}{2a}\\ =\frac{-6}{2\left(1\right)}\\ =3\end{array}$

The equation of the symmetry is $3$ then the x coordinate of the vertex is $3$

(b)

To determine

To Find: The table of the values that includes the vertex.

Answer to Problem 5.1.7EP

The required table is shown in Table 1.

Explanation of Solution

The given equation is $f\left(x\right)=x^2+6x-3$ .

The Table for the values that includes the vertex is shown in Table 1.

Table 1

$x$	$x^2+6x-3$	$f\left(x\right)$	$\left(x,f\left(x\right)\right)$
$-4$	${\left(-4\right)}^2+6\left(-4\right)-3$	$-11$	$\left(-4,-11\right)$
$-5$	${\left(-5\right)}^2+6\left(-5\right)-3$	$-8$	$\left(-5,-8\right)$
$-3$	${\left(-3\right)}^2+6\left(-3\right)-3$	$-12$	$\left(-3,-12\right)$
$0$	${\left(0\right)}^2+6\left(0\right)-3$	$-3$	$\left(0,-3\right)$
$1$	${\left(1\right)}^2+6\left(1\right)-3$	$4$	$\left(1,4\right)$

(c)

To determine

To Find: The graph for the function.

Answer to Problem 5.1.7EP

The required graph is shown in Figure 1.

Explanation of Solution

From the values of table 1 the graph is shown in Figure 1.

  

Figure 1