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2 years ago

The vertex of this parabola is at (-2, 5). Which of the following could be its equation?

Mathematics

1 answer:

LenKa [72]2 years ago

5 0

Answer:

y=3(x+2)^2+5

Step-by-step explanation:

The vertex is (p, q) where the "p" value is the x-coordinate of the vertex and the "q" is the y-coordinate of the vertex.

Given the vertex (-2, 5) the "p" here is -2 and "q" here is 5, so plugging into the vertex form we get y=3(x+2)^2+5

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Y = -x<br><br> What is is the constant of proportionality

Stels [109]

Hey!

------------------------------------------------

-x = -1

This statement shows that y equals -1. Since it doesn't show a table or word problem then we already know that the constant of proportionality is -1. Or if you just have the expression just divide the number by 1. -x/1 = -1!

------------------------------------------------

Thus, -1 is the constant of proportionality!

------------------------------------------------

Hope This Helped! Good Luck!

4 0

2 years ago

Let f(x)=5x3−60x+5 input the interval(s) on which f is increasing. (-inf,-2)u(2,inf) input the interval(s) on which f is decreas

o-na [289]

Answers:

(a) f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing at $(-2,2)$ .

(c) f is concave up at $(2, \infty)$

(d) f is concave down at $(-\infty, 2)$

Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that

$f'(x) = 15x^2 - 60
$

So,

$
f'(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \ 0} \text{ (1)}$

The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:

-->> x < -2
-->> -2 < x < 2
--->> x > 2

If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.

If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.

So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since

$f'(x) \ \textgreater \ 0 \Leftrightarrow (x - 2)(x + 2) \ \textgreater \ 0$

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing only when the derivative of f is negative. Since

$f'(x) = 15x^2 - 60$

Using the similar computation in (a),

$f'(x) \ \textless \ \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \ 0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \ 0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \ 0} \text{ (2)}$

Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.

Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

(c) f is concave up if and only if the second derivative of f is positive. Note that

$f''(x) = 30x - 60$

Since,

$f''(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 30x - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 30(x - 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow x - 2 \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{x \ \textgreater \ 2}$

Therefore, f is concave up at $(2, \infty)$ .

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,

$f''(x) = 30x - 60$

Using the similar computation in (c),

$f''(x) \ \textless \ 0 
\\ \\ \Leftrightarrow 30x - 60 \ \textless \ 0 
\\ \\ \Leftrightarrow 30(x - 2) \ \textless \ 0 
\\ \\ \Leftrightarrow x - 2 \ \textless \ 0 
\\ \\ \Leftrightarrow \boxed{x \ \textless \ 2}$

Therefore, f is concave down at $(-\infty, 2)$ .

3 0

3 years ago

A parabola has a minimum value of 0, a y-intercept of 4, and an axis of symmetry at x = -2

sveta [45]

Since the minimum value is 0 and axis of symmetry is -2 this means that the vertex is at -2,0 now with the y intercept of 4. You can now plug the values into Vertex form which will be y=a(x-h)^2+k. a being the shrink or stretch of the parabola, h being the x value of the vertex, and k being the y value of the vertex. with all of that plugged in it should look like y=(x+2)^2. You can check this equation by plugging in 0 as x which should find the y intercept of 4. So it should then look like y=(0+2)^2 -> y=(2)^2 -> y=4

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2 years ago

6) What is the Converse of the following

Sladkaya [172]

Answer:

B) If you don't do you homework, then you can't go to the movies

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2 years ago

A sample of seventh and eighth grade students were asked how many minutes they spent online last night. The results were recorde

Stolb23 [73]

Answer:

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3 years ago