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pickupchik [31]
3 years ago
8

What is the equation written in vertex from of a parabola with a vertex of (4, –2) that passes through (2, –14)?

Mathematics
1 answer:
vichka [17]3 years ago
8 0

Answer:

y = -3(x - 4)² - 2

Step-by-step explanation:

Given the vertex, (4, -2), and the point (2, -14):

We can use the vertex form of the quadratic equation:

y = a(x - h)² + k

Where:

(h, k) = vertex

a  =  determines whether the graph opens up or down, and it also makes the parent function <u>wider</u> or <u>narrower</u>.

  • <u>positive</u> value of a = opens <u><em>upward</em></u>
  • <u>negative</u> value of a = opens <u><em>downward</em></u>
  • a is between 0 and 1, (0 < a < 1) the graph is <u><em>wider</em></u> than the parent function.
  • a > 1, the graph is <u><em>narrower</em></u> than the parent function.

<em>h </em>=<em> </em>determines how far left or right the parent function is translated.

  • h = positive, the function is translated <em>h</em> units to the right.
  • h = negative, the function is translated |<em>h</em>| units to the left.

<em>k</em> determines how far up or down the parent function is translated.

  • k = positive: translate <em>k</em> units <u><em>up</em></u>.
  • k = negative, translate <em>k</em> units <u><em>down</em></u>.

Now that I've set up the definitions for each variable of the vertex form, we can determine the quadratic equation using the given vertex and the point:

vertex (h, k): (4, -2)

point (x, y): (2, -14)

Substitute these values into the vertex form to solve for a:

y = a(x - h)² + k

-14 = a(2 - 4)²  -2

-14 = a (-2)² -2

-14 = a4 + -2

Add to to both sides:

-14 + 2 = a4 + -2 + 2

-12 = 4a

Divide both sides by 4 to solve for a:

-12/4 = 4a/4

-3 = a

Therefore, the quadratic equation inI vertex form is:

y = -3(x - 4)² - 2

The parabola is downward-facing, and is vertically compressed by a factor of -3. The graph is also horizontally translated 4 units to the right, and vertically translated 2 units down.

Attached is a screenshot of the graph where it shows the vertex and the given point, using the vertex form that I came up with.

Please mark my answers as the Brainliest, if you find this helpful :)

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katovenus [111]

1. Let a and b be coefficients such that

\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}

Combining the fractions on the right gives

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2. a. The given ODE is separable as

x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}

Using the result of part (1), integrating both sides gives

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C

Given that y = 1 when x = 1, we find

\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)

so the particular solution to the ODE is

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)

We can solve this explicitly for y :

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)

\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|

\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|

\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}

2. b. When x = 9, we get

y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}

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