Chapter 4.2, Problem 8PPSE

Solution

To find: The vertex, axis of symmetry, maximum or minimum value and the range of parabola $y=x^2+2x+1$ .

The vertex is $\left(-1,0\right)$ , the axis of symmetry is the line $x=-1$ , the minimum value is $f\left(-\frac{b}{2a}\right)=0$ and the range of the parabola is $\left[0,\infty \right)$ .

Given:

The parabola is $y=x^2+2x+1$ . So, re-write the parabola as $y=f\left(x\right)$ :

  $f\left(x\right)=x^2+2x+1\mathrm{......}\left(1\right)$

Concept used:

The equation $f\left(x\right)=a{x}^2+bx+c$ where $a\ne 0$ of the quadratic function in standard form represents a parabola which has:

Vertex $\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)\mathrm{......}\left(2\right)$

The axis of symmetry is $x-$ coordinate of the vertex or the line $x=-\frac{b}{2a}\mathrm{......}\left(3\right)$

The graph of the function opens upwards if $a>0$ and open downwards if $a<0$ .

The function has maximum value $f\left(-\frac{b}{2a}\right)$ for $a<0$ at $x=-\frac{b}{2a}\mathrm{......}\left(4\right)$ .

The function has minimum value $f\left(-\frac{b}{2a}\right)$ for $a>0$ at $x=-\frac{b}{2a}\mathrm{......}\left(5\right)$ .

And if $a>0$ the range of the function is $\left[f\left(-\frac{b}{2a}\right),\infty \right)$ . For $a<0$ , the range of the function is $\left(-\infty ,f\left(-\frac{b}{2a}\right)\right]\mathrm{......}\left(6\right)$

Calculation:

Comparing the parabola $\left(1\right)$ with the standard form of quadratic function, $a=1,b=2\text{ and }c=1$ .

Evaluate for $f\left(-\frac{b}{2a}\right)$ ,

  $\begin{array}{c}f\left(-\frac{b}{2a}\right)=f\left(-\frac{2}{2\left(1\right)}\right)\\ =f\left(-1\right)\\ ={\left(-1\right)}^2+2\left(-1\right)+1\mathrm{....}\left(\because f\left(x\right)=x^2+2x+1\right)\\ =0\mathrm{......}\left(6\right)\end{array}$

Now with the help of $\left(2\right)$ , the vertex is evaluated as:

  $\begin{array}{c}\text{Vertex}\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)=\left(-\frac{\left(2\right)}{2\left(1\right)},f\left(-\frac{\left(2\right)}{2\left(1\right)}\right)\right)\mathrm{....}\left(a=1,b=2\text{ and }c=1\right)\\ =\left(-1,f\left(-1\right)\right)\\ =\left(-1,0\right)\mathrm{....}\left(\text{Using }\left(6\right),f\left(-1\right)=0\right)\end{array}$

Hence, the vertex is $\left(-1,0\right)$ .

With the help of $\left(2\right)$ , the axis of symmetry is the line $x=-1$ because the vertex is $\left(-1,0\right)$ .

Using $\left(5\right)$ , the function has minimum value $f\left(-\frac{b}{2a}\right)$ since $a>0$ . And from $\left(6\right)$ , the minimum value is $f\left(-\frac{b}{2a}\right)=0$ as evaluated. Therefore, the minimum value of the function is $f\left(-\frac{b}{2a}\right)=0$ .

Using $\left(6\right)$ , the range of the function for $a>0$ is $\left[f\left(-\frac{b}{2a}\right),\infty \right)\text{ i}\text{.e}\text{. }\left[0,\infty \right)$ .

Conclusion:

Hence, the vertex is $\left(-1,0\right)$ , the axis of symmetry is the line $x=-1$ , the minimum value is $f\left(-\frac{b}{2a}\right)=0$ and the range of the parabola is $\left[0,\infty \right)$ .