Answer:
x-intercepts are (0, 0) and (-6, 0)
Step-by-step explanation:
equation of a parabola in vertex form: y = a(x - h)² + k
where (h, k) is the vertex
Substituting the given vertex (-3, -18) into the equation:
y = a(x + 3)² - 18
If the y-intercept is (0, 0) then substitute x=0 and y=0 into the equation and solve for a:
0 = a(0 + 3)² - 18
⇒ 0 = a(3)² - 18
⇒ 0 = 9a - 18
⇒ 9a = 18
⇒ a = 2
Therefore, y = 2(x + 3)² - 18
To find the x-intercepts, set the equation to 0 and solve for x:
2(x + 3)² - 18 = 0
Add 18 to both sides: 2(x + 3)² = 18
Divide both sides by 2: (x + 3)² = 9
Square root both sides: x + 3 = ±3
Subtract 3 from both sides: x = ±3 - 3
so x = 3 - 3 = 0
and x = -3 - 3 = -6
So x-intercepts are (0, 0) and (-6, 0)
(5,41)
(-5,23)
(-10,14)
(-15,5)
Answer:
-2, 5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75, 82, 89, 96, 103
A=27.18
There is a photo showing the answer and the formula and stuff hope this helps
Answer:
Using either method, we obtain: 
Step-by-step explanation:
a) By evaluating the integral:
![\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cint%5Climits%5Et_0%20%7B%5Csqrt%5B8%5D%7Bu%5E3%7D%20%7D%20%5C%2C%20du)
The integral itself can be evaluated by writing the root and exponent of the variable u as: ![\sqrt[8]{u^3} =u^{\frac{3}{8}](https://tex.z-dn.net/?f=%5Csqrt%5B8%5D%7Bu%5E3%7D%20%3Du%5E%7B%5Cfrac%7B3%7D%7B8%7D)
Then, an antiderivative of this is: 
which evaluated between the limits of integration gives:

and now the derivative of this expression with respect to "t" is:

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:
"If f is continuous on [a,b] then

is continuous on [a,b], differentiable on (a,b) and 
Since this this function
is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:
