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sashaice [31]
2 years ago
7

On a coordinate plane, a curved line with a minimum value of (negative 1.25, negative 3.25) and a maximum value of (0.25, negati

ve 1.75), crosses the x-axis at (negative 2.25, 0), and crosses the y-axis at (0, negative 2). The line exits the plane at (negative 2.75, 6) and (1.5, 6).
Which statement is true about the end behavior of the graphed function?

As the x-values go to positive infinity, the function's values go to negative infinity.
As the x-values go to zero, the function's values go to positive infinity.
As the x-values go to negative infinity, the function's values are equal to zero.
As the x-values go to positive infinity, the function's values go to positive infinity.
Mathematics
1 answer:
Juliette [100K]2 years ago
8 0

The correct option is:

As the x-values go to positive infinity, the function's values go to positive infinity.

<h3>What is the end behavior of the function?</h3>

By the description, We know that the function has the points:

(-1.25, -3.25), (0.25, -1.75) (-2.25, 0), (0, -2), (-2.75, 6), (1.5, 6)

Notice that from x = 0.25 to x = 1.5 the function goes upwards, then in the right side, the function goes up, so as x goes to positive infinity, also does f(x).

On the left side, we can see  that from x = -2.75 to x = -2.25 the function goes downwards.

So, as x goes to the left (negative infinity) f(x) goes upwards.

Then as x tends to negative infinity, f(x) tends to infinity.

The correct option is:

As the x-values go to positive infinity, the function's values go to positive infinity.

If you want to learn more about end behavior:

brainly.com/question/1365136

#SPJ1

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Line segment 19 units long running from (x,0) ti (0, y) show the area of the triangle enclosed by the segment is largest when x=
Debora [2.8K]
The area of the triangle is

A = (xy)/2

Also,

sqrt(x^2 + y^2) = 19

We solve this for y.

x^2 + y^2 = 361

y^2 = 361 - x^2

y = sqrt(361 - x^2)

Now we substitute this expression for y in the area equation.

A = (1/2)(x)(sqrt(361 - x^2))

A = (1/2)(x)(361 - x^2)^(1/2)

We take the derivative of A with respect to x.

dA/dx = (1/2)[(x) * d/dx(361 - x^2)^(1/2) + (361 - x^2)^(1/2)]

dA/dx = (1/2)[(x) * (1/2)(361 - x^2)^(-1/2)(-2x) + (361 - x^2)^(1/2)]

dA/dx = (1/2)[(361 - x^2)^(-1/2)(-x^2) + (361 - x^2)^(1/2)]

dA/dx = (1/2)[(-x^2)/(361 - x^2)^(1/2) + (361 - x^2)/(361 - x^2)^(1/2)]

dA/dx = (1/2)[(-x^2 - x^2 + 361)/(361 - x^2)^(1/2)]

dA/dx = (-2x^2 + 361)/[2(361 - x^2)^(1/2)]

Now we set the derivative equal to zero.

(-2x^2 + 361)/[2(361 - x^2)^(1/2)] = 0

-2x^2 + 361 = 0

-2x^2 = -361

2x^2 = 361

x^2 = 361/2

x = 19/sqrt(2)

x^2 + y^2 = 361

(19/sqrt(2))^2 + y^2 = 361

361/2 + y^2 = 361

y^2 = 361/2

y = 19/sqrt(2)

We have maximum area at x = 19/sqrt(2) and y = 19/sqrt(2), or when x = y.
3 0
3 years ago
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