Question

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

Answer 1

Answer:

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)

Step-by-step 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:

$x=\pm\sqrt{12}=\pm2\sqrt{3}\text{ or } x=\pm\sqrt{-3}=\pm i\sqrt{3}$

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