Chapter 10.2, Problem 78AYU

(A)

To determine

To show:

The graph of the parabola is of the given form

Explanation of Solution

Given:

The equation of parabola

  $A{x}^2+Dx+Ey+F=0$

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

  $A{x}^2+Dx+Ey+F=0$

Vertical line if $E=0and{D}^2-4AF=0$

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

Vertical line if $E=0and{D}^2-4AF=0$

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

So the quadratic form is

  $D^2-4AF=0$

Therefore a vertical line

(C)

To determine

To show:

The graph of the parabola is of the given form

Explanation of Solution

Given:

The equation of parabola

  $A{x}^2+Dx+Ey+F=0$

Vertical line if $E=0and{D}^2-4AF>0$

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

Vertical line if $E=0and{D}^2-4AF>0$

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

So the quadratic form is

  $D^2-4AF>0$

Therefore two vertical line

(D)

To determine

To show:

The graph of the parabola is of the given form

Explanation of Solution

Given:

The equation of parabola

  $A{x}^2+Dx+Ey+F=0$

Vertical line if $E=0and{D}^2-4AF<0$

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

Vertical line if $E=0and{D}^2-4AF<0$

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

So the quadratic form is

  $D^2-4AF<0$

Therefore no pointsof the parabola