Q. 12 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 9–14, (a) graph each quadratic function by determining whether its graph opens up or down and by finding its vertex, axis of symmetry, y-intercept, and x-intercepts, if any. (b) Determine the domain and the range of the function. (c) Determine where the function is increasing and where it is decreasing.

$f\left(x\right)=9x^2+6x+1$

Step-by-Step Solution

Answer

Part (a) The graph is 

Part (b). Domain : $\left(-\infty ,\infty \right)$

              Range : $[0,\infty )$

Part (c). Decreasing : $\left(-\infty ,-\frac{1}{3}\right)$

              Increasing : $\left(-\frac{1}{3},\infty \right)$

Step 1. Given information.

The given function is $f\left(x\right)=9x^2+6x+1$.

Part (a) Step 1. Compare the function with standard form.

The standard form of quadratic function is $f\left(x\right)=a{x}^2+bx+c;a\ne 0$.

So, $a=9,b=6,c=1$

Now determine weather the parabola open up or down $\left(a<0, \mathrm{or} a>0\right)$.

Here $a>0$, so parabola open up. 

Part (a) Step 2. Determine the vertex.

The vertex is $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$.

So, $-\frac{b}{2a}=-\frac{6}{2·9}=-\frac{1}{3}$.

Also, $f\left(-\frac{b}{2a}\right)=f\left(-\frac{1}{3}\right)=9{\left(-\frac{1}{3}\right)}^2+6\left(-\frac{1}{3}\right)+1=1-2+1=0$.

Then, the vertex has coordinates$\left(-\frac{1}{3},0\right)$.

And the axis of symmetry is $x=-\frac{b}{2a}=-\frac{1}{3}$

Part (a) Step 3. Find x-intercept and y-intercept.

First check the discriminant and then find the x-intercept and y-intercept.

Discriminant: $D=b^2-4ac=6^2-4·9·1=36-36=0$

So, $D=0$ which means that the graph of the given function has one x-intercept.

To find x-intercept solve $f\left(x\right)=0 \mathrm{or} 9x^2+6x+1=0$.

  $$\begin{gathered} x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ {x}_{1,2}=\frac{-6\pm \sqrt{6^2-4·1·9}}{2·9} \\ {x}_{1,2}=\frac{-6\pm \sqrt{36-36}}{18} \\ {x}_{1,2}=\frac{-6}{18}=-\frac{1}{3} \end{gathered}$$

So, the x-intercept is $x=-\frac{1}{3}$.

And  y-intercept $f\left(0\right)=9{\left(0\right)}^2+6\left(0\right)+1=1$.

Part (a) Step 4. Graph the function.

The graph of the function is  

Part (b) Step 1. Determine the domain and the range of the function.

The domain of function is the set of all real numbers. Based on the graph, the range of function is the interval $[0,\infty )$.

Part (c) Step 1. Determine where the function is increasing and where it is decreasing.

The function f is decreasing on the interval $\left(-\infty ,-\frac{1}{3}\right)$ and is increasing on the interval $\left(-\frac{1}{3},\infty \right)$.