Question

An equation and its graph are given. Find the x- and y-intercepts.

$x^4+y^2-xy=16$

Short answer

The x-intercepts are and 2 and y-intercepts are and 4.

Step-by-step solution

Step 1. Apply the concept of intercepts.

The x-intercepts are the points on x-axis where the graph of the equation intersects the x-axis.

The y-intercepts are the points on the y-axis where the graph of the equation intersects the y-axis.

Step 2. Compute the x-intercept.

Consider the provided equation$x^4+y^2-xy=16$ and the graph of the equation is sketched below.

Observe from the graph that the curve intersects thex-axis at two points and 2.

Recall that the x-intercepts are the points on the x-axis where the graph of the equation intersects the x-axis.

Substitute $y=0$in the equation $x^4+y^2-xy=16$,

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

Either$x=-2$ or $x=2$.

Therefore, x-intercepts are and 2.

Step 3. Compute the y-intercept.

Consider the provided equation$x^4+y^2-xy=16$ and the graph of the equation is sketched below.

Observe from the graph that the curve intersects they-axis at two points $-4$and 4.

Recall that the y-intercepts are the points on the y-axis where the graph of the equation intersects the y-axis.

Substitute $x=0$in the equation $x^4+y^2-xy=16$.

$$\begin{gathered} 0^4+y^2-x\left(0\right)=16 \\ {y}^2=16 \end{gathered}$$

Either $y=-4$or $y=4$.

Therefore, x-intercepts are $-4$and 4.

Thus, the x-intercepts are$-2$ and 2 and y-intercepts are$-4$ and 4.