Q. 1 - Precalculus Enhanced with Graphing Utilities

Question

The intercepts of the equation $x^2+4y^2=16$ are ______________________________. 

Step-by-Step Solution

Answer

The intercepts are $\left(4, 0\right), \left(-4, 0\right), \left(0, 2\right) \mathrm{and} \left(0,-2\right)$.

Step 1. Given information

The given equation is $x^2+4y^2=16$.

Need to find the intercepts of the given equation.

Step 2. Finding y-intercept

Substitute $x=0$ in the equation $x^2+4y^2=16$ and solve for y to obtain the y-intercepts.

$\begin{array}{rcl}{\left(0\right)}^2+4{\left(y\right)}^2& =& 16\\ 4y^2& =& 16\\ y^2& =& 4\\ y^2-4& =& 0\\ \left(y-2\right)\left(y+2\right)& =& 0\end{array}$

$y=2 or y=-2$

The intercepts on the y-axis are $\left(0, 2\right)$ and $\left(0,-2\right)$.

Step 3. Finding x-intercept

Substitute $y=0$ in the equation $x^2+4y^2=16$ and solve for$x$ to obtain the x-intercepts.

$\begin{array}{rcl}{x}^2+4{\left(0\right)}^2& =& 16\\ x^2& =& 16\\ x^2-16& =& 0\\ \left(x-4\right)\left(x+4\right)& =& 0\end{array}$

$\begin{array}{rcl}x& =& 4 or x=-4\end{array}$

The intercepts on the x-axis are $\left(4, 0\right)$ and $\left(-4, 0\right)$.