Question

PLEASE HELP!

Identify the y-intercept and zeros of
y = x^4 - 9x^2 - 36
Algebraically​

Answer 1

Our y-intercept is (0, -36).

Our real zeros are: (-2√3, 0) and (2√3, 0)

And our imaginary zeros are: (i√3, 0) and (-i√3, 0)

Explanation:

We have the function:

`y=x^4-9x^2-36`

Let's identify the y- and x-intercepts.

Y-Intercept:

To determine the y-intercept, we will substitute 0 for x and solve for y. So:

`y=(0)^4-9(0)^2-36`

Evaluate:

`y=-36`

Therefore, our y-intercept is at (0, -36).

X-Intercepts:

To determine our x-intercepts, we need to substitute 0 for y and solve for x. So:

`0=x^4-9x^2-36`

Now, we can factor. Before doing so, we can make the factored easier to see by converting this into quadratic form. Let's let u=x². Then:

`0=(u^2)^2-9(x^2)-36`

Substitute:

`0=u^2-9u-36`

Now, let's factor. We can use -12 and 3. So:

`0=u^2+3u-12u-36 \\ 0=(u(u+3)-12(u+3)) \\ 0=(u-12)(u+3)`

Now, we can substitute back u:

`0=(x^2-12)(x^2+3)`

Zero Product Property:

`x^2-12=0\text{ or } x^2+3=0`

Solve for x in each case:

`x^2=12 \text{ or } x^2=-3`

Take the square root of both sides. Since we're taking an even root, we will need plus/minus. Therefore:

`x=\pm√(12)=\pm2√(3)\text{ or } x=\pm√(-3)=\pm i√(3)`

Therefore, our zeros (both real and imaginary) are: (2√3, 0), (-2√3, 0), (i√3, 0), and (-i√3, 0)