Answer:
y=5(x-3)^2+2
Step-by-step explanation:
You plug the point into f(x)=a(x-3)^2+2... that already has already been solved for the vertex that you want. Then you swap it out for the solution you have solved for.
well from me studing this question i have found the My KeY bOaRd JuSt BrOkE
This is honestly simple but we all need help! The answer is 55,053. Hope this helps! =3
Answer:
The answer is 0
Step-by-step explanation:
To comlete each square, you just have to take the middle term and divide it by 2, then take its square. Add this to both sides of the equation then simplify. The middle term is the 1st degree x variable.
<span>3. x^2+6x =16
-------------------------
(6/2)^2 = 9
x</span>² + 6x + 9 = 16 + 9
(x + 3)² = 25
<span>4. x^2-10x =11
------------------------
(-10/2)</span>² = 25
x² - 10x + 25 = 11 + 25
(x -5)² = 36
<span>5. x^2-9x =0
--------------
(-9/2)</span>² = 81/4
x² - 9x + 81/4 = 81/4
(x - 9/2)² = 81/4
<span>6. x^2+16x =15
---------------
(16/2)</span>² = 64
x² + 16x +64 = 15 + 64
(x + 8)² = 79
<span>7. 3x^2+18x-81=0
----------------
x</span>² + 6x = 27
(6/2)² = 9
x² + 6x + 9 = 27 + 9
(x + 3)² = 36
A polynomial with a general equation of ax² + bx + c = 0 has a root determined using this formula:

Just substitute the coefficients to the formula to find the roots
8. 1.09, -1.84
9. 3, 1/2
10. No real roots
11. 1.12, -3.12