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4 years ago

In two or more complete sentences, describe how you would draw the graph of the solution set for the following inequality.

Mathematics

2 answers:

Eddi Din [679]4 years ago

8 0

Answer:

Step-by-step explanation:

podryga [215]4 years ago

7 0

A nonlinear function that can be written on the standard form

ax2+bx+c,wherea≠0ax2+bx+c,wherea≠0

is called a quadratic function.

All quadratic functions has a U-shaped graph called a parabola. The parent quadratic function is

y=x2y=x2

The lowest or the highest point on a parabola is called the vertex. The vertex has the x-coordinate

x=−b2ax=−b2a

The y-coordinate of the vertex is the maximum or minimum value of the function.

a > 0 parabola opens up minimum value

a < 0 parabola opens down maximum value

A rule of thumb reminds us that when we have a positive symbol before x2we get a happy expression on the graph and a negative symbol renders a sad expression.

The vertical line that passes through the vertex and divides the parabola in two is called the axis of symmetry. The axis of symmetry has the equation

x=−b2ax=−b2a

The y-intercept of the equation is c.

When you want to graph a quadratic function you begin by making a table of values for some values of your function and then plot those values in a coordinate plane and draw a smooth curve through the points.

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4 years ago

Help please 50 points!

uranmaximum [27]

Answer:

A and B

Step-by-step explanation:

we would like to solve the following equation:

$\rm\displaystyle 5 - 2x = \sqrt{ {2x}^{2} + x - 1 } - x$

to do so isolate -x to the left hand side and change its sign:

$\rm\displaystyle 5 - 2x + x = \sqrt{ {2x}^{2} + x - 1 }$

simplify addition:

$\rm\displaystyle 5 - x = \sqrt{ {2x}^{2} + x - 1 }$

square both sides:

$\rm\displaystyle (5 - x {)}^{2} = (\sqrt{ {2x}^{2} + x - 1 } {)}^{2}$

simplify square of the right hand side:

$\rm\displaystyle (5 - x {)}^{2} = {2x}^{2} + x - 1$

use (a-b)²=a²-2ab+b² to expand the left hand side:

$\rm\displaystyle {x}^{2} - 10x + 25= {2x}^{2} + x - 1$

swap the equation:

$\rm\displaystyle {2x}^{2} + x - 1 = {x}^{2} - 10x + 25$

isolate the right hand side expression to the left hand side and change every sign:

$\rm\displaystyle {2x}^{2} + x - 1 - {x}^{2} + 10x - 25 = 0$

simplify:

$\rm\displaystyle {x}^{2} + 11x - 26= 0$

rewrite the middle term as 13x-2x:

$\rm\displaystyle {x}^{2} + 13x - 2x - 26= 0$

factor out x:

$\rm\displaystyle x({x}^{} + 13)- 2x - 26= 0$

factor out -2:

$\rm\displaystyle x({x}^{} + 13)- 2(x + 13)= 0$

group:

$\rm\displaystyle (x- 2)(x + 13)= 0$

by Zero product property we obtain:

$\displaystyle \begin{cases}x- 2 = 0\\ x + 13= 0 \end{cases}$

solve for x:

$\displaystyle \begin{cases}x = 2\\ x = - 13 \end{cases}$

to check for extraneous solutions we can define the domain of equation recall that a square root of a function is always greater than or equal to 0 therefore

$\rm\displaystyle 5 - x \geq0$