Answer :
[tex]y=\frac{1}{2}x^2-4\\\\zeros\ when\ y=0\\\\\frac{1}{2}x^2-4=0\\\\\frac{1}{2}x^2=4\ \ \ \ /\cdot2\\\\x^2=8\\\\x=-\sqrt8\ and\ x=\sqrt8\\\\\underline{\underline{x\approx-2.8\ and\ x\approx2.8}}[/tex]
Answer:
x= 2.8 and x= -2.8
Step-by-step explanation:
[tex]y=\frac{1}{2}x^2-4[/tex]
To find out the zeros, we replace x y with 0 and solve for x
[tex]0=\frac{1}{2}x^2-4[/tex]
Add 4 on both sides
[tex]4=\frac{1}{2}x^2[/tex]
Multiply by 2 on both sides
8= x^2
Now we take square root on both sides
x= +- [tex]\sqrt{8}[/tex]
x=2.82843 and x=-2.82843
x= 2.8 and x= -2.8
