Joshua buys five bottles of orange juice she has coupons for $.65 off the regular price of each bottle of juice after using the
2 answers:
You might be interested in
Answer:
A
Step-by-step explanation:
So we have the quadratic equation and it's written in three equivalent forms:
Let's determine the characteristics of the quadratic equation with the given equations.
From the first equation, since the leading coefficient (0.5) is positive, we can be certain that the graph opens upwards.
Also, the constant term is -1.5, so the y-intercept is (0,-1.5).
The second equation is the vertex form. Vertex form has the format:
Where (h,k) is the vertex. From the second equation we know that h is -1 (because (x+1) is the same as (x-(-1))) and k is -2. Therefore, the vertex is (-1,-2).
And since the graph points upwards, this means that (-1,-2) is the minimum point of the function. In other words, the range of the function is greater than or equal to -2. In interval notation, this is:
This also means that the end behavior of the graph as a x approaches negative and positive infinity is positive infinity because the graph will always go straight up.
Also, the third form is the factored form. With that, we can solve for the zeros of the quadratic. The zeros are:
Therefore, the graph crosses the x-axis at x=-3 and x=1.
So, from the three equations, we gathered the following information:
1) The graph curves upwards.
2) The roots of zeros of the function is (-3,0) and (1,0).
3) The y-intercept is (0,-1.5).
4) The vertex is (-1,-2). This is also the minimum point.
5) Therefore, the range of the graph is all values greater than or equal to -2.
6) The end behavior of the graph on both directions go towards positive infinity.
Therefore, our correct answer is A.
B is not correct because the line of symmetry (or the x-coordinate of the vertex) here is -1 and not 1/2.
C is not correct because the graph goes towards <em>positive </em>infinity since it shoots straight up.
And D is not correct because the y-intercept is (0,-1.5).
I suppose you mean
Recall that
which converges everywhere. Then by substitution,
which also converges everywhere (and we can confirm this via the ratio test, for instance).
a. Differentiating the Taylor series gives
(starting at because the summand is 0 when
)
b. Naturally, the differentiated series represents
To see this, recalling the series for , we know
Multiplying by gives
and from here,
c. This series also converges everywhere. By the ratio test, the series converges if
The limit is 0, so any choice of satisfies the convergence condition.