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Oksanka [162]
2 years ago
8

The figure below shows a partial graph of two functions, f(x)f(x) and g(x)g(x). f(x)\f(x) is a parabola defined by the quadratic

function f(x)=-(x+2)^{2}+17f(x)=−(x+2)
2
+17 and g(x)g(x) is a linear function that passes through the points \left(0,-1\right)(0,−1) and (1,0)(1,0).


Respond to Part A and Part B based on the given information.
​
Part A
Determine the x-x−coordinate of the point in Quadrant III where \ f(x)=g(x) f(x)=g(x).




Part B
Justify algebraically why the xx-coordinate you identified in Part A is a solution to the equation f(x)=g(x).f(x)=g(x).
Mathematics
1 answer:
Marina86 [1]2 years ago
4 0

The x-coordinate of the point in quadrant III where f(x) = g(x) is -7

<h3>How to determine the x-coordinate in quadrant III?</h3>

The figure is not attached to the question; however, the question can still be solved

The quadratic function is given as:

f(x) = -(x +2)² + 17

The linear function passes through (0,-1) and (1,0).

So, we calculate the linear equation using:

y = \frac{y_2 - y_1}{x_2 -x_1} * (x -x_1) + y_1

Substitute known values

y = \frac{0+ 1}{1-0} * (x -0) -1

Evaluate

y = x - 1

So, we have:

  • Quadratic function: f(x) = (x +2)² + 17
  • Linear function: g(x) = x - 1

Next, we plot the graph of both functions (see attachment)

From the attached graph, both functions meet in the quadrant III at (-7,-8)

Hence, the x-coordinate of the point in quadrant III where f(x) = g(x) is -7

<h3>Justify the solution in (a), algebraically </h3>

In (a), we have the solution to be:

(x,y) = (-7,-8)

Substitute -7 for x in f(x) and g(x)

f(-7) = -(-7 +2)² + 17 = -8

g(-7) = -7 - 1 = -8

See that:

f(-7) = g(-7) = -8 and it is located in the third quadrant.

Hence, the solution in (a) is true

Read more about equation at:

brainly.com/question/2972832

#SPJ1

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a. Write in terms of x the number of tickets for children
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  substitute equation 1 to equation 2.
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Tap for fewer steps...

Factor out the GCF of  

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from each term in the polynomial.

Tap for fewer steps...

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from the expression  

63

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21

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(

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+

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.

21

b

2

c

2

(

3

c

2

)

+

21

b

2

c

2

(

2

b

)

Since all the terms share a common factor of  

21

b

2

c

2

, it can be factored out of each term.

21

b

2

c

2

(

3

c

2

+

2

b

)

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