Question

find the x- and y-intercepts.
\\(\frac{(x + 6)^2}{9} + \frac{(y - 4)^2}{4} = 1\\)
write each intercept as an ordered pair.
if there is more than one intercept, use the \and\ button.
select
one\ if applicable.
part 1 of 2
x-intercept(s): \\(\boxed{}\\) \\(\frac{\square}{\square}\\) \\((\square,\square)\\) \\(\square\text{and}\square\\) \\(\text{none}\\)

Explanation:

Step1: Find x-intercepts, set $y=0$

Substitute $y=0$ into the equation:
$$\frac{(x+6)^2}{9} + \frac{(0-4)^2}{4} = 1$$

Step2: Simplify the constant term

$$\frac{(x+6)^2}{9} + \frac{16}{4} = 1 \implies \frac{(x+6)^2}{9} + 4 = 1$$

Step3: Isolate the x-term

$$\frac{(x+6)^2}{9} = 1 - 4 = -3$$

Step4: Analyze the equation

Since $(x+6)^2 \geq 0$ for all real $x$, $\frac{(x+6)^2}{9} \geq 0$, which cannot equal $-3$. So no real x-intercepts.

Step5: Find y-intercepts, set $x=0$

Substitute $x=0$ into the equation:
$$\frac{(0+6)^2}{9} + \frac{(y-4)^2}{4} = 1$$

Step6: Simplify the constant term

$$\frac{36}{9} + \frac{(y-4)^2}{4} = 1 \implies 4 + \frac{(y-4)^2}{4} = 1$$

Step7: Isolate the y-term

$$\frac{(y-4)^2}{4} = 1 - 4 = -3$$

Step8: Analyze the equation

Since $(y-4)^2 \geq 0$ for all real $y$, $\frac{(y-4)^2}{4} \geq 0$, which cannot equal $-3$. So no real y-intercepts.

Answer:

x-intercept(s): None
y-intercept(s): None