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Expert-verified Found in: Page 252 ### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508 # A quadratic function $\mathbit{f}$ is given.(a) Express $\mathbit{f}$ in standard form.(b) Find the vertex and $\mathbit{x}$- and $\mathbit{y}$- intercepts of $\mathbit{f}$.(c) Sketch a graph of $\mathbit{f}$.(d) Find the domain and range of $\mathbit{f}$.$\mathbit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{3}{\mathbit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbit{x}\mathbf{-}\mathbf{2}$

(a) The equation of $f$ in standard form is $\mathbit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{3}{\mathbf{\left(}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}\mathbf{1}$.

(b) The vertex is $\mathbf{\left(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{\right)}$ and the $x$-intercepts are $\frac{\mathbf{3}\mathbf{+}\sqrt{\mathbf{3}}}{\mathbf{3}}$ and $\frac{\mathbf{3}\mathbf{-}\sqrt{\mathbf{3}}}{\mathbf{3}}$ and the $y$ intercept is $\mathbf{-}\mathbf{2}$.

(c) The graph of $f$ is: (d) The domain and range of $f$ are $\mathbf{\left(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{\right)}$ and $\mathbf{\left(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{1}\mathbf{\right]}$ respectively.

See the step by step solution

## Part a Step 1. Write the definition of standard form of quadratic function.

The standard form of the quadratic function $f\left(x\right)=a{x}^{2}+bx+c$ is given by:

$f\left(x\right)=a{\left(x-h\right)}^{2}+k$.

Where $\left(h,k\right)$ is the vertex of the quadratic function $f\left(x\right)=a{x}^{2}+bx+c$.

## Part a Step 2. Express f in standard form.

The given quadratic function is $f\left(x\right)=-3{x}^{2}+6x-2$.

Convert the given quadratic function $f\left(x\right)=-3{x}^{2}+6x-2$ in standard form by using the method of completing square.

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

Therefore, the standard form of the given quadratic function $f\left(x\right)=-3{x}^{2}+6x-2$ is $f\left(x\right)=-3{\left(x-1\right)}^{2}+1$.

## Part b Step 1. Find the values of a, h and k.

The given quadratic function in standard form is $f\left(x\right)=-3{\left(x-1\right)}^{2}+1$.

Compare the quadratic function $f\left(x\right)=-3{\left(x-1\right)}^{2}+1$ with $f\left(x\right)=a{\left(x-h\right)}^{2}+k$ to find the values of $a,h$ and $k$.

Therefore, it is obtained that:

$a=-3,h=1$ and $k=1$

Therefore, the vertex of the quadratic function $f\left(x\right)=-3{\left(x-1\right)}^{2}+1$ is given by:

$\left(h,k\right)=\left(1,1\right)$.

## Part b Step 2. Find the x- and y- intercepts of f.

To find the $x$-intercept substitute 0 for $f\left(x\right)$ in the equation $f\left(x\right)=-3{\left(x-1\right)}^{2}+1$.

Therefore, it is obtained that:

$\begin{array}{l}f\left(x\right)=-3{\left(x-1\right)}^{2}+1\\ 0=-3{\left(x-1\right)}^{2}+1\\ 3{\left(x-1\right)}^{2}=1\\ {\left(x-1\right)}^{2}=\frac{1}{3}\\ x-1=±\sqrt{\frac{1}{3}}\\ x=1±\frac{1}{\sqrt{3}}\\ x=1±\frac{\sqrt{3}}{3}\\ x=\frac{3±\sqrt{3}}{3}\\ x=\frac{3+\sqrt{3}}{3}\text{or}x=\frac{3-\sqrt{3}}{3}\end{array}$

Therefore, the $x$-intercepts are $\frac{3+\sqrt{3}}{3}$ and $\frac{3-\sqrt{3}}{3}$.

To find the $y$-intercept substitute 0 for $x$ in the equation role="math" localid="1648535359672" $f\left(x\right)=-3{\left(x-1\right)}^{2}+1$.

Therefore, it is obtained that:

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

Therefore, the $y$-intercept is $-2$.

## Part c Step 1. Draw the rough sketch of the quadratic function f(x)=a(x−h)2+k when a<0.

The rough sketch of the quadratic function $f\left(x\right)=a{\left(x-h\right)}^{2}+k$ when $a<0$ is given by: ## Part c Step 2. Sketch a graph of f.

The graph of $f$ is: ## Part d Step 1. Write the definition of domain and range of the function.

The Domain of the function is the set of the values of the independent variable for which the function is defined.

The Range of the function is set of the values of the dependent variable which are obtained by substituting the values of the independent variable which are in the domain of the function.

## Part d Step 2. Find the domain and range of f.

From the given function $f\left(x\right)=-3{x}^{2}+6x-2$, it can be noticed that $x$ is the independent variable and $y$ is the dependent variable.

Therefore, the domain of $f$ is the set of values of $x$ for which the function is defined and range is set of the values of $y$ which are obtained by substituting the values of $x$ which are in the domain of the function.

From the graph of the function $f\left(x\right)=-3{x}^{2}+6x-2$, it can be noticed that $x\in \left(-\infty ,\infty \right)$ and $y\in \left(-\infty ,1\right]$.

Therefore, the Domain and range of the function $f$ are $\left(-\infty ,\infty \right)$ and $\left(-\infty ,1\right]$ respectively. ### Want to see more solutions like these? 