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:
[tex]y=x^4-9x^2-36[/tex]
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:
[tex]y=(0)^4-9(0)^2-36[/tex]
Evaluate:
[tex]y=-36[/tex]
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:
[tex]0=x^4-9x^2-36[/tex]
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:
[tex]0=(u^2)^2-9(x^2)-36[/tex]
Substitute:
[tex]0=u^2-9u-36[/tex]
Now, let's factor. We can use -12 and 3. So:
[tex]0=u^2+3u-12u-36 \\ 0=(u(u+3)-12(u+3)) \\ 0=(u-12)(u+3)[/tex]
Now, we can substitute back u:
[tex]0=(x^2-12)(x^2+3)[/tex]
Zero Product Property:
[tex]x^2-12=0\text{ or } x^2+3=0[/tex]
Solve for x in each case:
[tex]x^2=12 \text{ or } x^2=-3[/tex]
Take the square root of both sides. Since we're taking an even root, we will need plus/minus. Therefore:
[tex]x=\pm\sqrt{12}=\pm2\sqrt{3}\text{ or } x=\pm\sqrt{-3}=\pm i\sqrt{3}[/tex]
Therefore, our zeros (both real and imaginary) are: (2√3, 0), (-2√3, 0), (i√3, 0), and (-i√3, 0)
