Find the range of the parent function below y=x^2

1 answer:

6 0

you don't show the graph, but graphing y=x^2 it is a U shaped line that starts at (0,0) and the lines go upwards on both sides of the Y axis,

This means all real numbers above 0 and 0 will solve it

so the range is y≥0

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in two or more complete sentences, describe how to graph the following equation. in your description, include the slope and both

garik1379 [7]

Answer:

Slope: 0

x-intercept: there isnt one

y-intercept: 2

Step-by-step explanation:

the easiest way to look at this is to put it in the form y=mx+b where m is the slope and b is the y intercept. when we just think about what y=2 would look like, we can imagine a straight horizontal line at y=2. No matter what x value you choose, y will always equal 2. We know the slope (m) of any horizontal line is zero because there is no rise and zero divided by anything is going to be zero. we also know if y is always equal to 2 the y intercept will be 2. this would give us y=0x+2. to find the x intercept we just need to set y equal to zero in this equation. this gives us 0=0x+2 or 0=2 which can never be true, therefore there will be no x intercept.

5 0

3 years ago

The diameter of Circle terminates on the circumference of the circle at (3,0) and (3,−7).. Write the equation of the circle in

Verdich [7]

Step-by-step explanation:

that means the center of the circle is exactly in the middle between the 2 points.

and since both points have the same x value (so, the diameter here is a vertical line), it is ready to get the y value of the middle : (-7 - 0) / 2 = -3.5

so, the coordinates of the center are (3, -3.5).

and that makes the standard circle equation

(x - h)² + (y - k)² = r²

(x - 3)² + (y + 3.5)² = 3.5² = 12.25

as h, k are the coordinates of the center, and r is the radius (half the diameter).

7 0

2 years ago

Answer the question with explanation;

PSYCHO15rus [73]

Answer:

The statement in the question is wrong. The series actually diverges.

Step-by-step explanation:

We compute

$\lim_{n\to\infty}\frac{n^2}{(n+1)^2}=\lim_{n\to\infty}\left(\frac{n^2}{n^2+2n+1}\cdot\frac{1/n^2}{1/n^2}\right)=\lim_{n\to\infty}\frac1{1+2/n+1/n^2}=\frac1{1+0+0}=1\ne0$

Therefore, by the series divergence test, the series $\sum_{n=1}^\infty\frac{n^2}{(n+1)^2}$ diverges.

EDIT: To VectorFundament120, if $(x_n)_{n\in\mathbb N}$ is a sequence, both $\lim x_n$ and $\lim_{n\to\infty}x_n$ are common notation for its limit. The former is not wrong but I have switched to the latter if that helps.