Chapter 9, Problem 1GA

(a)

To determine

The x- and y- intercepts using analytical method

Answer to Problem 1GA

Solution:x-intercepts are

$\left(5,0\right)\text{and}\left(-5,0\right)$

y-intercepts are

$\left(0,5\right)\text{and}\left(0,-5\right)$

Explanation of Solution

Given:

  $x^2+y^2=25$

Concept Used:

• x-intercept is the point on the graph where the value of y is zero.
• y-intercept is the point on the graph where the value of x is zero.
• Square rot property:- If $x^2=y$ , then $x=\pm \sqrt{y}$

Calculation:

We have been given the equation $x^2+y^2=25$. In order to find the x-intercept, we plug $y=0$ in the given equation and solve for x

Therefore, x-intercepts are

$\left(5,0\right)\text{and}\left(-5,0\right)$

Now, for y-intercepts we plug

$x=0$ in the given equation and solve for y

  $\begin{array}{l}{x}^2+y^2=25\left(\text{Given equation}\right)\\ \Rightarrow 0^2+y^2=25\left(\text{Plug }x=0\right)\\ \Rightarrow y^2=25\left(\text{Simplify}\right)\\ \Rightarrow \sqrt{y^2}=\pm \sqrt{25}\left(\text{Apply square root property}\right)\\ \Rightarrow y=\pm 5\left(\text{Simplify }\right)\end{array}$

Therefore, y-intercepts are

$\left(0,5\right)\text{and}\left(0,-5\right)$

Conclusion:

x-intercepts are

$\left(5,0\right)\text{and}\left(-5,0\right)$

y-intercepts are

$\left(0,5\right)\text{and}\left(0,-5\right)$

(b)

To determine

To solve: The equation for y

Answer to Problem 1GA

Solution:

  $\begin{array}{l}{Y}_1=-\sqrt{25-x^2}\\ Y_2=\sqrt{25-x^2}\end{array}$

Explanation of Solution

Given:

  $x^2+y^2=25$

Concept Used:

Square rot property:- If

$x^2=y$ , then

$x=\pm \sqrt{y}$

Calculation:

We have been given the equation $x^2+y^2=25$ and we have to solve it for y. In order to solve for y, we will do below mentioned steps.

  $\begin{array}{l}{x}^2+y^2=25\left(\text{Given equation}\right)\\ \Rightarrow y^2=25-x^2\left(\text{Subtract }{x}^2\text{ to both sides}\right)\\ \Rightarrow \sqrt{y^2}=\pm \sqrt{25-x^2}\left(\text{Apply square root property}\right)\\ \Rightarrow y=\pm \sqrt{25-x^2}\left(\text{Simplify}\right)\\ \Rightarrow y=-\sqrt{25-x^2},\sqrt{25-x^2}\end{array}$

Therefore, we have

  $\begin{array}{l}{Y}_1=-\sqrt{25-x^2}\\ Y_2=\sqrt{25-x^2}\end{array}$

Conclusion:

Using square root property, we have

  $\begin{array}{l}{Y}_1=-\sqrt{25-x^2}\\ Y_2=\sqrt{25-x^2}\end{array}$

(c)

To determine

To graph: The two functions from part (b) on a graphing calculator. Use a window

$-7.6\le x\le 7.6,-5\le y\le 5$

Answer to Problem 1GA

Solution:

Explanation of Solution

Given:

  $\begin{array}{l}{Y}_1=-\sqrt{25-x^2}\\ Y_2=\sqrt{25-x^2}\end{array}$

Calculation:

From part (b), we have the two functions as

  $\begin{array}{l}{Y}_1=-\sqrt{25-x^2}\\ Y_2=\sqrt{25-x^2}\end{array}$

Using a graphing calculator, below is the graph of the functions in the window

$-7.6\le x\le 7.6,-5\le y\le 5$

(d)

To determine

If the functions graphed on the calculator intersect with the x - axis. Why do the x - intercepts not show up in the graph? If you relied solely on your calculator to find the x -intercepts, do you think would always get reliable results?

Explanation of Solution

No, the functions graphed on the calculator do not seem to intersect with the x - axis.

Since, we have split the original equation in two functions and these two functions intersect at the x −intercepts of the original equation. Hence, the x − intercepts do not show up in the graph.

No, we can't rely solely on the graphing calculator for finding x −intercepts. This is because the calculator has limited resolution.