In the parabola y=(x+1)^2+2

Mathematics

1 answer:

Svet_ta [14]3 years ago

3 0

Direction: Opens Up

Vertex: (1,2)

Focus: (1, 9/4)

Axis of Symmetry: x = 1

Directrix: y = 7/4

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1) What is the equation of the line whose y-intercept is 3 and slope is 1?

s2008m [1.1K]

$The\ slope-intercept\ form:y=mx+b\\\\m\ is\ slope\\b\ is\ y-intercept\\\\m=1;\ b=3\ then\ y=1x+3\Rightarrow y=x+3$

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2 years ago

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Consider the following equations. f(x) = − 4/ x3, y = 0, x = −2, x = −1. Sketch the region bounded by the graphs of the equation

Sindrei [870]

Answer:

1.5 unit^2

Step-by-step explanation:

Solution:-

- A graphing utility was used to plot the following equations:

$f ( x ) = - \frac{4}{x^3}\\\\y = 0 , x = -1 , x = -2$

- The plot is given in the document attached.

- We are to determine the area bounded by the above function f ( x ) subjected boundary equations ( y = 0 , x = -1 , x = - 2 ).

- We will utilize the double integral formulations to determine the area bounded by f ( x ) and boundary equations.

We will first perform integration in the y-direction ( dy ) which has a lower bounded of ( a = y = 0 ) and an upper bound of the function ( b = f ( x ) ) itself. Next we will proceed by integrating with respect to ( dx ) with lower limit defined by the boundary equation ( c = x = -2 ) and upper bound ( d = x = - 1 ).

The double integration formulation can be written as:

$A= \int\limits_c^d \int\limits_a^b {} \, dy.dx \\\\A = \int\limits_c^d { - \frac{4}{x^3} } . dx\\\\A = \frac{2}{x^2} |\limits_-_2^-^1\\\\A = \frac{2}{1} - \frac{2}{4} \\\\A = \frac{3}{2} unit^2$