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LenKa [72]
3 years ago
12

Create the quadratic function that contains the points (0-2), (1, 0) and (3, 10). Show all of your work for full credit.

Mathematics
1 answer:
solniwko [45]3 years ago
8 0

Answer:

My graphing calculator tells me the function is

.. f(x) = x^2 +x -2

_____

Since we recognize the y-intercept as being (0, -2), we only need to find the coefficients of x and x^2.

If you want to do this by hand, you can use

.. y = ax^2 +bx -2

to write 2 equations in a and b for the (x, y) values (1, 0) and (3, 10).

.. 0 = a +b -2

.. 10 = 9a +3b -2

To solve by substitution, you can let a = 2-b, and you have

.. 10 = 9(2 -b) +3b -2

.. 6b = 6 . . . . . . . . . . . add 6b-10

.. b = 1

.. a = 2 -1 = 1

Step-by-step explanation:

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3 0
3 years ago
Read 2 more answers
Find the horizontal and vertical asymptotes of​ f(x). ​f(x) equals = StartFraction 6 x Over x plus 2 EndFraction 6x x+2 Find the
miss Akunina [59]

Answer:

The horizontal asymptote can be described by the line y = 6

The vertical asymptote can be described by the line x = -2

Step-by-step explanation:

* <em>Lets the meaning of vertical and horizontal asymptotes</em>

- <u><em>Vertical asymptotes</em></u> are vertical lines which correspond to the zeroes

  of the denominator of a rational function

- <u><em>A horizontal asymptote</em></u> is a y-value on a graph which a function

 approaches but does not actually reach

- If the degree of the numerator is less than the degree of the

 denominator, then there is a horizontal asymptote at y = 0

- If the degree of the numerator is greater than the degree of the

 denominator, then there is no horizontal asymptote

- If the degree of the numerator is equal the degree of the denominator,

 then there is a horizontal asymptote at y = leading coefficient of the

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* <em>Lets solve the problem</em>

∵ f(x)=\frac{6x}{x+2}

∵ The numerator is 6x

∵ The denominator is x + 2

∴ The numerator and the denominator have same degree

∵ The leading coefficient of the numerator is 6

∵ The leading coefficient of the denominator is 1

∴ There is a horizontal asymptote at y = 6/1

∴ <em>The horizontal asymptote can be described by the line y = 6</em>

- Put the denominator equal zero to find its zeroes

∵ The denominator is x + 2

∴ x + 2 = 0

- Subtract 2 from both sides

∴ x = -2

∴ <em>The vertical asymptote can be described by the line x = -2</em>

6 0
3 years ago
Read 2 more answers
How do you solve 3x+7=x
vredina [299]
3x + 7 = x

First, subtract 3x from both sides. / Your problem should look like: 7 = x - 3x
Second, simplify x - 3x to -2x. / Your problem should look like: 7 = -2x
Third, divide both sides by -2. / Your problem should look like: \frac{7}{-2} = x
Fourth, simplify \frac{7}{-2} to -\frac{7}{2} / Your problem should look like: - \frac{7}{2} = x
Fifth, switch sides. / Your problem should look like: x = -\frac{7}{2}

Answer as fraction: -\frac{7}{2}
Answer as decimal: -3.5


8 0
3 years ago
Read 2 more answers
Can someone help me solve #9!?
yarga [219]
Dy/dx = (4y²)(x⁴/³)

Find ∫(4y²)(x⁴/³) =∫(4y²∛(x⁴)dx = 3∛(x⁴).y² +c or 3x⁴/³.y² + c
4 0
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