PreCalculus - 4.1.1

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

The intercepts of the equation \(\displaystyle 9x^2 + 4y = 36\) are \(\displaystyle \cdots\)

 

[mathdad]

Let me see if I remember. The intercept of an equation is the location on the xy-plane where the graph crosses the x-axis, y-axis or both. Yes?

To find the y-intercept, let x = 0 and solve for y. To find the x-intercept, let y = 0 and solve for x.

Let x = 0

9(0)^2 + 4y = 36

0 + 4y = 36

4y = 36

4y/4 = 36/4

y = 9

The y-intercept is 9 located at the point (0, 9).
Right?

Let y = 0

9x^2 + 4(0) = 36

9x^2 + 0 = 36

9x^2 = 36

9(x^2)/9 = 36/9

x^2 = 4

sqrt{x^2} = sqrt{4}

x = -2, x = 2

The x-intercepts are x = -2 and x = 2. We find x = -2 at the point (-2, 0) and x = 2 at the point (2, 0).

Right?

 

[logistic_guy]

Let me see if I remember. The intercept of an equation is the location on the xy-plane where the graph crosses the x-axis, y-axis or both. Yes?

To find the y-intercept, let x = 0 and solve for y. To find the x-intercept, let y = 0 and solve for x.

Let x = 0

9(0)^2 + 4y = 36

0 + 4y = 36

4y = 36

4y/4 = 36/4

y = 9

The y-intercept is 9 located at the point (0, 9).
Right?

Let y = 0

9x^2 + 4(0) = 36

9x^2 + 0 = 36

9x^2 = 36

9(x^2)/9 = 36/9

x^2 = 4

sqrt{x^2} = sqrt{4}

x = -2, x = 2

The x-intercepts are x = -2 and x = 2. We find x = -2 at the point (-2, 0) and x = 2 at the point (2, 0).

Right?

\(\displaystyle \textcolor{blue}{\bold{Correct.}}\)