Chapter 3.CT, Problem 1CT

To determine

To Graph:

The quadratic function $f(x)=-2x^2+6x-3$ and give the x-intercept, vertex, axis, domain, range and the largest open intervals of the domain over which the function is increasing and decreasing.

Answer to Problem 1CT

Solution:

The x-intercept is $\frac{3+\sqrt{3}}{2}and\frac{3-\sqrt{3}}{2}$ or $2.366and0.634$.

The y-intercept is $-3$.

The vertex of the quadratic function is $\left(\frac{3}{2},\frac{3}{2}\right)\iff \left(1.5,1.5\right)$.

The axis of symmetry is $\frac{3}{2}$.

The domain of the function is $\left(-\infty ,\infty \right)$ and the range of the function is $(-\infty ,1.5)$.

The function is decreasing from the interval $(1.5,\infty )$ and increasing from the interval $(-\infty ,1.5)$.

The graph of the quadratic function is shown as below:

Explanation of Solution

Given:

The provided quadratic function is $f(x)=-2x^2+6x-3$.

Concept Used:

Vertex

If the quadratic function is of the form $a{x}^2+bx+c$ then the vertex of the quadratic function is $\left(-\frac{b}{2a},f\left(-\frac{b}{2a}\right)\right)$.

X-intercept

The x-intercept is where a line crosses the x-axis.

Y-intercept

The y-intercept is where a line crosses the y-axis.

Axis of Symmetry:

The axis of symmetry is $x=-\frac{b}{2a}$.

Domain:

The domain of the quadratic function is $\left(-\infty ,\infty \right)$

Calculation:

X-intercept

Substitute the value of $y=0$ in an quadratic function $-2x^2+6x-3$, we get

$\begin{array}{l}0=-2x^2+6x-3\\ 2x^2-6x+3=0\end{array}$

Compare the above equation with general quadratic equation $a{x}^2+bx+c=0$, we get

$a=2,b=-6\text{and}c=3$

Use the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, we get

$\begin{array}{rll}x& =& \frac{-(-6)\pm \sqrt{{(-6)}^2-4(2)(3)}}{2\cdot 2}\\ & =& \frac{6\pm \sqrt{36-24}}{4}\\ & =& \frac{6\pm \sqrt{12}}{4}\\ & =& \frac{3\pm \sqrt{3}}{2}\end{array}$

The x-intercept is $\frac{3+\sqrt{3}}{2}and\frac{3-\sqrt{3}}{2}$ or $2.366and0.634$.

Y-intercept

Substitute the value of $x=0$ in an quadratic function $-2x^2+6x-3$, we get

$\begin{array}{l}y=-2{(0)}^2+6(0)-3\\ y=-3\end{array}$

The y-intercept is $-3$.

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

$\begin{array}{rll}x& =& -\frac{b}{2a}\\ & =& -\frac{-6}{2\times 2}\\ & =& \frac{3}{2}\end{array}$

$\begin{array}{rll}f\left(\frac{3}{2}\right)& =& -2\cdot {\left(\frac{3}{2}\right)}^2+6\cdot \frac{3}{2}-3\\ & =& \frac{3}{2}\end{array}$

Therefore, the vertex of the quadratic function is $\left(\frac{3}{2},\frac{3}{2}\right)\iff \left(1.5,1.5\right)$.

The axis of symmetry is $\frac{3}{2}$.

$x$	$f(x)$
1	$-2{(1)}^2+6(1)-3=1$
2	$-2{(2)}^2+6(2)-3=1$
0.5	$-2{(0.5)}^2+6(0.5)-3=-0.5$
2.5	$-2{(2.5)}^2+6(2.5)-3=-0.5$

Graph:

From the graph, we can determine the domain and range of the function.

The domain of the function is $\left(-\infty ,\infty \right)$ and the range of the function is $(-\infty ,1.5)$.

The function is decreasing from the interval $(1.5,\infty )$ and increasing from the interval $(-\infty ,1.5)$.

Final Statement:

The x-intercept is $\frac{3+\sqrt{3}}{2}and\frac{3-\sqrt{3}}{2}$ or $2.366and0.634$.

The y-intercept is $-3$.

The vertex of the quadratic function is $\left(\frac{3}{2},\frac{3}{2}\right)\iff \left(1.5,1.5\right)$.

The axis of symmetry is $\frac{3}{2}$.

The domain of the function is $\left(-\infty ,\infty \right)$ and the range of the function is $(-\infty ,1.5)$.

The function is decreasing from the interval $(1.5,\infty )$ and increasing from the interval $(-\infty ,1.5)$.