The function f(x)=x^2+5x+6 has zeros at -2 and -3. What are the zeros of the function y=-2f(x/11) ?
Do you just plug x/11 into each x value in f(x)? Like f(x)= (x/11)^2+5(x/11)+6 (I exclueded the -2 because if I divide 0 by -2, it's still zero)?
But then how can you find the zeros? Factor?
if f(x)=x^2+5x+6
then
-2f(x/11) = -2( (x/11)^2 + 5(x/11) + 6)
you want the zeros of that ...
-2[x^2/121 + 5x/11 + 6] = 0
x^2/121 + 5x/11 + 6 = 0 , I think you had it that far
times 121
x^2 + 55x + 726 = 0
(x+22)(x+33) = 0
x = -22 or x = -33
To find the zeros of the function y = -2f(x/11), we need to substitute x/11 in place of x in the original function f(x) = x^2 + 5x + 6.
So, let's compute f(x/11):
f(x/11) = (x/11)^2 + 5(x/11) + 6
Now, we can simplify this expression:
f(x/11) = x^2/121 + 5x/11 + 6
Next, we multiply the entire function by -2:
y = -2(x^2/121 + 5x/11 + 6)
= -2x^2/121 - 10x/11 - 12
Now, we have the new function y = -2f(x/11). To find the zeros of this function, we set y equal to zero and solve for x:
0 = -2x^2/121 - 10x/11 - 12
To simplify the equation, we can multiply every term by 121 to get rid of the denominator:
0 = -2x^2 - 1110x - 1452
Now, we can attempt to factorize the quadratic equation, but it doesn't appear to be easily factorable. In such cases, we can use the quadratic formula to find the zeros:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -2, b = -1110, and c = -1452. Substituting these values into the quadratic formula, we get:
x = (-(-1110) ± √((-1110)^2 - 4(-2)(-1452))) / (2(-2))
x = (1110 ± √(1232100 - 11616)) / (-4)
x = (1110 ± √(1221484)) / (-4)
x = (1110 ± 1106.78) / (-4)
Using these values, we can calculate the two zeros of the function y = -2f(x/11):
x₁ = (1110 + 1106.78) / (-4)
= 549.44 / (-4)
= -137.36
x₂ = (1110 - 1106.78) / (-4)
= 3.22 / (-4)
= -0.805
So, the zeros of the function y = -2f(x/11) are approximately -137.36 and -0.805.