Answer:
4(x-7)^2 - 19
Step-by-step explanation:
Shifting 7 units to the right: 4(x-7)^2 - 16.
3 units down: 4(x-7)^2 - 16 - 3, or 4(x-7)^2 - 19
Quartic is 4th degree
the factors of an equation with roots r1,r2 is
(x-r1)(x-r2)
4th degree
it could be
(x-r1)¹(x-r2)³ or
(x-r1)²(x-r2)² or
(x-r1)³(x-r2)¹
roots or zeroes at x=-1 and x=-2
(x-(-1)) and (x-(-2))
(x+1) and (x+2)
the function could be factored into
(x+1)¹(x+2)³ or
(x+1)²(x+2)² or
(x+1)³(x+2)¹
expanded would be
x⁴+7x³+18x²+20x+9 or
x⁴+6x³+13x²+12x+4 or
x⁴+5x³+9x²+7x+2
one of those is the answer
Answer: 15 Units
Step-by-step explanation:
Im smart
So it tells us that g(3) = -5 and g'(x) = x^2 + 7.
So g(3) = -5 is the point (3, -5)
Using linear approximation
g(2.99) is the point (2.99, g(3) + g'(3)*(2.99-3))
now we just need to simplify that
(2.99, -5 + (16)*(-.01)) which is (2.99, -5 + -.16) which is (2.99, -5.16)
So g(2.99) = -5.16
Doing the same thing for the other g(3.01)
(3.01, g(3) + g'(3)*(3.01-3))
(3.01, -5 + 16*.01) which is (3.01, -4.84)
So g(3.01) = -4.84
So we have our linear approximation for the two.
If you wanted to, you could check your answer by finding g(x). Since you know g'(x), take the antiderivative and we will get
g(x) = 1/3x^3 + 7x + C
Since we know g(3) = -5, we can use that to solve for C
1/3(3)^3 + 7(3) + C = -5 and we find that C = -35
so that means g(x) = (x^3)/3 + 7x - 35
So just to check our linear approximations use that to find g(2.99) and g(3.01)
g(2.99) = -5.1597
g(3.01) = -4.8397
So as you can see, using the linear approximation we got our answers as
g(2.99) = -5.16
g(3.01) = -4.84
which are both really close to the actual answer. Not a bad method if you ever need to use it.
Answer:
hope this helps
Step-by-step explanation:
1 i believe because its the only one that says "more than". The other two say less than.
