Chapter 10.2, Problem 79AYU

(A)

To determine

To show:

The graph of the parabola is of the given form

Explanation of Solution

Given:

The equation of parabola

  $C{x}^2+Dx+Ey+F=0$

Concept used:

The stabdard form of parabola equation

  $y=a{x}^2+bx+c$

Putting $x=h,y=k$ in equation (1)

  $k=a{h}^2+bh+c\mathrm{.......................}\left(2\right)$

Where $\left(h,k\right)$ is the vertex

Calculation:

The stabdard form of parabola equation

  $y=a{x}^2+bx+c$

Putting $x=h,y=k$ in equation (1)

  $\begin{array}{l}k=a{h}^2+bh+c\\ {\left(y-k\right)}^2=4a\left(x-h\right)\mathrm{.......................}\left(2\right)\end{array}$

Where $\left(h,k\right)$ is the vertex

Draw the table

  $A{x}^2+Dx+Ey+F=0$

Test one point in each of the region formed by the graph

If the point satisfies the function then shade the entire region to denote that every point in the region satisfies the function

$x-axis$	$-0.5$	$0$	$0.65$	$-1$
$y-axis$	$-0.75$	$-1$	$-2$	$-1$

  

(B)

To determine

To show:

The graph of the parabola is of the given form

Explanation of Solution

Given:

The equation of parabola

  $C{x}^2+Dx+Ey+F=0$

Horizontal line if $D=0and{E}^2-4CF=0$

Concept used:

The stabdard form of parabola equation

  $y=a{x}^2+bx+c$

Putting $x=h,y=k$ in equation (1)

  $k=a{h}^2+bh+c\mathrm{.......................}\left(2\right)$

Where $\left(h,k\right)$ is the vertex

Calculation:

The stabdard form of parabola equation

  $y=a{x}^2+bx+c$

Putting $x=h,y=k$ in equation (1)

  $\begin{array}{l}k=a{h}^2+bh+c\\ {\left(y-k\right)}^2=4a\left(x-h\right)\mathrm{.......................}\left(2\right)\end{array}$

Where $\left(h,k\right)$ is the vertex

Horizontal line if $D=0and{E}^2-4CF=0$

  $\begin{array}{l}C{x}^2+Dx+Ey+F=0\\ C{x}^2+0+Ey+F=0\\ C{x}^2+Ey+F=0\end{array}$

So the quadratic form is

  $E^2-4CF=0$

Therefore a horizental line

(C)

To determine

To show:

The graph of the parabola is of the given form

Explanation of Solution

Given:

The equation of parabola

  $C{x}^2+Dx+Ey+F=0$

Horizontal line if $D=0and{E}^2-4CF>0$

Concept used:

The stabdard form of parabola equation

  $y=a{x}^2+bx+c$

Putting $x=h,y=k$ in equation (1)

  $k=a{h}^2+bh+c\mathrm{.......................}\left(2\right)$

Where $\left(h,k\right)$ is the vertex

Calculation:

The stabdard form of parabola equation

  $y=a{x}^2+bx+c$

Putting $x=h,y=k$ in equation (1)

  $\begin{array}{l}k=a{h}^2+bh+c\\ {\left(y-k\right)}^2=4a\left(x-h\right)\mathrm{.......................}\left(2\right)\end{array}$

Where $\left(h,k\right)$ is the vertex

Horizontal line if $D=0and{E}^2-4CF>0$

  $\begin{array}{l}A{x}^2+Dx+Ey+F=0\\ A{x}^2+0+Ey+F=0\\ A{x}^2+Ey+F=0\end{array}$

So the quadratic form is

  $E^2-4CF>0$

Therefore two horizentzalline

(D)

To determine

To show:

The graph of the parabola is of the given form

Explanation of Solution

Given:

The equation of parabola

  $C{x}^2+Dx+Ey+F=0$

Horizontal line if $D=0and{E}^2-4CF<0$

Concept used:

The stabdard form of parabola equation

  $y=a{x}^2+bx+c$

Putting $x=h,y=k$ in equation (1)

  $k=a{h}^2+bh+c\mathrm{.......................}\left(2\right)$

Where $\left(h,k\right)$ is the vertex

Calculation:

The stabdard form of parabola equation

  $y=a{x}^2+bx+c$

Putting $x=h,y=k$ in equation (1)

  $\begin{array}{l}k=a{h}^2+bh+c\\ {\left(y-k\right)}^2=4a\left(x-h\right)\mathrm{.......................}\left(2\right)\end{array}$

Where $\left(h,k\right)$ is the vertex

Horizontal line if $D=0and{E}^2-4CF<0$

  $\begin{array}{l}C{x}^2+Dx+Ey+F=0\\ C{x}^2+Ey+0+F=0\\ C{x}^2+Ey+F=0\end{array}$

So the quadratic form is

  $E^2-4CF<0$

Therefore no points of the parabola