Well first you would find a common denominator which would be 1w and then since you multiplied 4 by 3 to get to 12 you multiply 3 by 3 which is 9. So then you would have 9/12 + 1/12 which = 10/12

3 0

2 years ago

i promise i will mark brainliest if the point (-3,4) is the vertex point of the curve of aquadratic function f and -5 is aroot o

Jet001 [13]

Ans:

(x+5)(x+1)+8

Step by step explanation:

-5 was mentioned as the 0 of the first root: (x+5). Then with some maneuvering we can plot (-3,4) then add (x+1) because they land on the same y axis. then we add 8 to get it right on the vertex.

6 0

3 years ago

Find the area of the region enclosed by f(x) and the x axis for the given function over the specified u reevaluate f(x)=x^2+3x+4

lord [1]

Answer:

A = 68 unit^2

Step-by-step explanation:

Given:-

The piece-wise function f(x) is defined over an interval as follows:

f(x) = { x^2+3x+4 , x < 3

f(x) = { x^2+3x+4 , x≥3

Domain : [ -3 , 4 ]

Find:-

Find the area of the region enclosed by f(x) and the x axis

Solution:-

- The best way to tackle problems relating to piece-wise functions is to solve for each part individually and then combine the results.

- The first portion of function is valid over the interval [ -3 , 3 ]:

$f(x) = x^2+3x+4$

- The area "A1" bounded by f(x) is given as:

$A1 = \int\limits^a_b {f(x)} \, dx$

Where, The interval of the function { -3 , 3 ] = [ a , b ]:

$A1 = \int\limits^a_b {x^2+3x+4} \, dx\\\\A1 = \frac{x^3}{3} + \frac{3x^2}{2} + 4x |\limits_-_3^3 \\\\A1 = \frac{3^3}{3} + \frac{3*3^2}{2} + 4*3 - \frac{-3^3}{3} - \frac{3(-3)^2}{2} - 4(-3)\\\\A1 = 9 + 13.5 +12 + 9-13.5+12\\\\A1 =42 unit^2$

- Similarly for the other portion of piece-wise function covering the interval [3 , 4] :

$f(x) = 4x+10$

- The area "A2" bounded by f(x) is given as:

$A2 = \int\limits^a_b {f(x)} \, dx$

Where, The interval of the function { 3 , 4 ] = [ a , b ]:

$A2 = \int\limits^a_b {4x+10} \, dx\\\\A2 = 2x^2 + 10x |\limits_3^4 \\\\A2 = 2*(4)^2 + 10*4 - 2*(3)^2 - 10*3\\\\A2 = 32 + 40 - 18-30\\\\A2 =26 unit^2$

- The total area "A" bounded by the piece-wise function over the entire domain [ -3 , 4 ] is given:

A = A1 + A2

A = 42 + 26

A = 68 unit^2

8 0

2 years ago