All categories

2 years ago

The function f is given in three equivalent forms. Which form most quickly reveals the zeros (or 'roots") of the function?

Mathematics

1 answer:

shepuryov [24]2 years ago

3 0

9514 1404 393

Answer:

A, 2

Step-by-step explanation:

In the factored form, the zeros are the values of x that make a factor zero. That is, one of the zeros is ...

x -2 = 0 ⇒ x = 2

To find the zeros from the other forms, more work must be done. In the standard form, one of the usual solution methods must be used, possibly including putting the equation into factored form, completing the square, or using the quadratic formula. In vertex form, a root must be taken, then those values adjusted by translation to the axis of symmetry.

You might be interested in

Decide which measurement is greater. Place the correct symbol ( <, >, =) on the line between the measurements.

laila [671]

Answer:

5>18 and 25

1 and 4>32

3<8

13<4

4<9

i hope this helped :)

Step-by-step explanation:

5 0

1 year ago

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

2 years ago

Aiden spins the spinner shown 20 times. His results are shown in the table. What are the experimental and theoretical probabilit

harkovskaia [24]

Answer:D

Step-by-step explanation:

5 0

2 years ago

Please answer I will give branliest

Leni [432]

Answer:

Step-by-step explanation:

assuming that weekends are only sat and sunday

24 students

if were assuming that friday is included

30 students

hope this helps <3

6 0

2 years ago

Read 2 more answers

-4/3 divided by 12/5

Ksivusya [100]

Answer:

-5/9

Step-by-step explanation:

−4/5 divided by 12 /5

=-20/36

simplify

=-5/9

5 0

2 years ago

Read 2 more answers