The parabola x = y² - 9 opens:

2 answers:

7 0

The parabola x = y² - 9 opens to the right. The vertex is at (-9,0). Since y has no negative coefficient, the parabola opens to the right. Parabola is two-dimensional and is a mirror-symmetrical curve. It is more or less U-shaped.

bogdanovich [222]3 years ago

3 0

Answer:

The parabola opens right.

Step-by-step explanation:

The given parabola is $x=y^2-9$

We know that:-

If the x term is square then the parabola opens upward or downward based on the value of a. If a>0 then parabola opens upward and if a<0 then the parabola open downwards.

If the y term is square then the parabola opens left or right based on the value of a. If a>0 then parabola opens right and if a<0 then the parabola open left.

Now, in the given equation, the y term is square. Hence, the parabola either opens left or right.

Now, comparing with $x=ay^2+by+c$

The value of a is 1 >0

Hence, the parabola open right.

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Show that if X is a geometric random variable with parameter p, then

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Answer:

$\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}$

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

In order to find the expected value E(1/X) we need to find this sum:

$E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}$

Lets consider the following series:

$\sum_{k=1}^{\infty} b^{k-1}$

And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:

$\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}$ (a)

On the last step we assume that $0\leq r\leq b$ and $\sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}$ , then the integral on the left part of equation (a) would be 1. And we have: