All categories

3 years ago

What is the graph of the function f(x)=x2+ 2x+ 3?

Mathematics

1 answer:

8090 [49]3 years ago

8 0

Answer:

Step-by-step explanation:

The best way to do this is to let a graphing program do it. You could do it from a chart, but a graphing program can be useful for that as well. We'll make up a mini chart here

x method y

-3 (-3)^2 + 2(-3) + 3 = 6

-2 (-2)^2 + 2(-2) + 3 = 3

-1 (-1)^2 + 2(-1) + 3 = 2

0 3

1 (1)^2 + 2(1) + 3 6

2 (2^2) + 2(2) + 3 11

3 (3)^2 +2(3) + 3 18

See the graph below.

You can pick it out from the graphs you were given.

You might be interested in

I need help ASAP no links and no tricks the one who answers correctly I will mark brainiest. If I got #5 wrong let me know and t

GaryK [48]

Answer:

$y=-\frac{4}{3} x+4$

Step-by-step explanation:

Slope-intercept form is written in $y=mx+b$

So, to solve, we must separate y onto one side and make its coefficient one.

$4x+3y=12$

Subtract 4x on both sides.

$3y=12-4x$

Divide both sides by the coefficient of $y$ , which is 3.

$\frac{3y}{3} =\frac{12-4x}{3}$

Simplify.

$y=4-\frac{4}{3} x$

Rewrite this equation into standard form by changing the order of $mx$ and $b$

$y=-\frac{4}{3} x+4$

3 0

2 years ago

Read 2 more answers

The number of minutes Pam talked on the phone was 35, 40, 12, 16, What was the average time Pam was on the phone

Fudgin [204]

Answer:

Step-by-step explanation:

(35 + 40 + 12 + 16) : 4

= (75 + 28) : 4

= 103 : 4

= 25,75

8 0

3 years ago

Which pair of points has a positive slope?

viva [34]

B.) (1,1) (2,5) should be your answer.

5 0

3 years ago

Read 2 more answers

Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

3 0

3 years ago

A(3,4) and B(-3,2) are pointd on a coordinate plane. find the coordinate of a points C on the x axis such that AC=BC

mr Goodwill [35]

Answer:

Step-by-step explanation:

Here's the game plan. In order to find a point on the x-axis that makes AC = BC, we need to find the midpoint of AB and the slope of AB. From there, we can find the equation of the line that is perpendicular to AB so we can then fit a 0 in for y and solve for x. This final coordinate will be the answer you're looking for. First and foremost, the midpoint of AB:

and

Now for the slope of AB:

and

So if the slope of AB is 1/3, then the slope of a line perpendicular to that line is -3. What we are finding now is the equation of the line perpendicular to AB and going through (0, 3):

and filling in:

y - 3 = -3(x - 0) and

y - 3 = -3x + 0 and

y - 3 = -3x so

y = -3x + 3. Filling in a 0 for y will give us the coordinate we want for the x-intercept (the point where this line goes through the x-axis):

0 = -3x + 3 and

-3 = -3x so

x = 1

The coordinate on the x-axis such that AC = BC is (1, 0)

7 0

2 years ago