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1 year ago

find the coordinates of the vertex of the following parabola algebraically. write your answer as an (x,y) point..y=x²+9

Mathematics

1 answer:

Alexandra [31]1 year ago

3 0

Explanation:

using the parabola formula:

y = a(x-h)² + k²

vertex = (h, k)

We are given a parebola equation of: y = x²+9

comparing both equations to get the vertex:

y = y

a = 1

(x-h)² = x²

x² = (x + 0)²

(x-h)² = (x + 0)²

h = 0

+k = +9

k = 9

The vertex of the parabola as (x, y): (0, 9)

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Twice as many students are in the chess club as in the fencing club.The number of students in the fencing club is one fifth the

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Answer:

28 students

Step-by-step explanation:

Let number of students in cooking club = x

Number of students in fencing club = 1/5 x

Number of students in chess club = 2(1/5)x

Since, There are 42 more students in the cooking club than in the chess club

Then ;

Number of students in chess club = Number of students in cooking club - 42

2(1/5)x = x - 42

(2/5)x - x = - 42

-0.6x = - 42

x = 42 / 0.6

x = 70

Hence, number of students in chess club :

(2/5) * 70 = 28

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Step-by-step explanation:

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What is an equation for the linear function whose graph contains the points (9, 7) and (4, −8) ?

aalyn [17]

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3 years ago

Find the taylor polynomial t3(x) for the function f centered at the number

inysia [295]

Answer:

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

Step-by-step explanation:

We are given that

$f(x)=7tan^{-1}(x)$

a=1

$T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}$

Substitute n=3 and a=1

$t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}$

$f(x)=7tan^{-1}(x)$

$f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}$

Where $tan^{-1}(1)=\frac{\pi}{4}$

$f'(x)=\frac{7}{1+x^2}$

Using the formula

$\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}$

$f'(1)=\frac{7}{2}$

$f''(x)=\frac{-14x}{(1+x^2)^2}$

$f''(1)=-\frac{7}{2}$

$f''(x)=-14x(x^2+1)^{-2}$

$f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})$

By using the formula

$(uv)'=u'v+v'u$

$f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}$

$f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}$

$f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}$

Substitute the values

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3$

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

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