Answer: [tex]y=-\frac{1}{4} x^2-4x-18[/tex]
Step-by-step explanation:
Hey there! Looks like you want to convert the vertex form of a quadratic into the standard form. This just means that you have to evaluate the function until there are no other operations other than addition and subtraction(shown below).
Standard form quadratic: [tex]\boxed{ax^2 + bx + c}[/tex]
Let's convert now!
Solving:
[tex]y=-\frac{1}{4} (x+8)^2-2[/tex]
Square [tex]x+8[/tex]:
[tex]y=-\frac{1}{4} (x^2+16x+64)-2[/tex]
Now multiply [tex](x^2+16x+64)[/tex] by [tex]-\frac{1}{4}[/tex]:
[tex]y=(-\frac{1}{4} x^2-4x-16)-2[/tex]
Combine like terms(-2 with -16):
[tex]\boxed{\boxed{y=-\frac{1}{4} x^2-4x-18}}[/tex]
That's it!
