Chapter 13, Problem 2CR

To determine

To find: The graph is upward or downward and find the axis of symmetry and intercept.

Answer to Problem 2CR

The parabola is open upward, the vertex $\left(-2,9\right)$ , axis of symmetry $x=-2$ , intercepts $\left(1,0\right),\left(-5,0\right)\text{ and }\left(0,-5\right)$ .

Explanation of Solution

Given information: $f\left(x\right)=x^2+4x-5$

Calculation: We know that if the leading coefficient is getter than zero then parabola opens upward if less than it’s open downward.

So, clearly the parabola open upward.

Now, the vertex form of a quadratic function is,

  $f\left(r\right)=a{\left(x-h\right)}^2+k,\text{ where,}\mathrm{........}\text{(I)}$

  $\left(h,k\right)$ is the vertex of the parabola.

  $x=h$ is the axis of symmetry.

Use completing the square method to convert $f\left(x\right)$ into vertex form.

  $f\left(x\right)=x^2+4x-5$ comparing by $a{x}^2+bx+c=0$

  $a=1,b=4,c=-5$

Now,

  $\begin{array}{c}{x}^2+4x-5=0\\ x^2+4x=5\\ x^2+4x+{\left(\frac{4}{2}\right)}^2=5+{\left(\frac{4}{2}\right)}^2\\ x^2+4x+4=5+4\\ {\left(x+2\right)}^2=9\\ {\left(x+2\right)}^2-9=0\end{array}$

So, $f\left(x\right)={\left(x+2\right)}^2-\mathrm{9......}(II)$

Comparing $\left(II\right)\text{ by (I)}$ then,

  $a=1,h=-2,k=9$

Here, The vertex is $\left(2,9\right)$ and axis of symmetry

  $\begin{array}{l}x=h\\ x=-2\end{array}$

X-intercept,

  $\begin{array}{c}{\left(x+2\right)}^2=9\\ x+2=\pm 3\\ x=1,-5\end{array}$

Hence, The X-intercepts is $\left(1,0\right)and\left(-5,0\right)$

Y-intercept,

Let, $x=0$

Now,

  $\begin{array}{c}f\left(x\right)={\left(x+2\right)}^2-9\\ f\left(0\right)=-5\end{array}$

Hence, The Y-intercept is $\left(0,5\right)$

Hence, The parabola is open upward, the vertex $\left(-2,9\right)$ , axis of symmetry $x=-2$ , intercepts $\left(1,0\right),\left(-5,0\right)\text{ and }\left(0,-5\right)$