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3 years ago

What is the standard form? Given the vertex (-3,3) and the focus point (-3,2)

Mathematics

1 answer:

antoniya [11.8K]3 years ago

5 0

Answer:

$\large\boxed{y=-\dfrac{1}{4}x^2-\dfrac{3}{2}x+\dfrac{3}{4}}$

Step-by-step explanation:

The equation of a parabola in the vertex form:

$y=a(x-h)^2+k$

(h, k) - vertex

Then the focus is:

$\left(h,\ k+\dfrac{1}{4a}\right)$

We have:

$the\ vertex\ (-3,\ 3)\to h=-3,\ k=3\\\\the\ focus\ (-3,\ 2)\to h=-3,\ k+\dfrac{1}{4a}=2$

Substitute k = 3 and calculate a:

$3+\dfrac{1}{4a}=2\qquad\text{subtract 3 from both sides}\\\\\dfrac{1}{4a}=-1\\\\\dfrac{1}{4a}=\dfrac{-1}{1}\qquad\text{cross multiply}\\\\(-1)(4a)=(1)(1)\\\\-4a=1\qquad\text{divide both sides by (-4)}\\\\a=-\dfrac{1}{4}$

Substitute to the vertex form:

$y=-\dfrac{1}{4}(x-(-3))^2+3$

Convert ot the standard form

$y=ax^2+bx+c\\\\y=-\dfrac{1}{4}(x+3)^2+3\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\y=-\dfrac{1}{4}(x^2+2(x)(3)+3^2)+3\\\\y=-\dfrac{1}{4}(x^2+6x+9)+3\qquad\text{use the distributive property}\\\\y=\left(-\dfrac{1}{4}\right)(x^2)+\left(-\dfrac{1}{4}\right)(6x)+\left(-\dfrac{1}{4}\right)(9)+3\\\\y=-\dfrac{1}{4}x^2-\dfrac{6}{4}x-\dfrac{9}{4}+\dfrac{12}{4}\\\\y=-\dfrac{1}{4}x^2-\dfrac{3}{2}x+\dfrac{3}{4}$

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(N * 3) + 6 - (n * 2) = n + 3 what is value of n

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3 years ago

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John runs a computer software store. Yesterday he counted 123 people who walked by the store, 56 of whom came into the store. Of

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Answer:

a) There is a 45.53% probability that a person who walks by the store will enter the store.

b) There is a 41.07% probability that a person who walks into the store will buy something.

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Step-by-step explanation:

This a probability problem.

The probability formula is given by:

$P = \frac{D}{T}$

In which P is the probability, D is the number of desired outcomes and T is the number of total outcomes.

The problem states that:

123 people walked by the store.

56 people came into the store.

23 bought something in the store.

(a) Estimate the probability that a person who walks by the store will enter the store.

123 people walked by the store and 56 entered the store, so $T = 123, D = 56$ .

$P = \frac{D}{T} = \frac{56}{123} = 0.4553$

There is a 45.53% probability that a person who walks by the store will enter the store.

(b) Estimate the probability that a person who walks into the store will buy something.

56 people came into the store and 23 bought something, so $T = 56, D = 23$ .

$P = \frac{D}{T} = \frac{23}{56} = 0.4107$

There is a 41.07% probability that a person who walks into the store will buy something.

(c) Estimate the probability that a person who walks by the store will come in and buy something.

123 people walked by the store and 23 came in and bought something, so $T = 123, D = 23$ .

$P = \frac{D}{T} = \frac{23}{123} = 0.1870$

There is a 18.70% probability that a person who walks by the store will come in and buy something.

(d) Estimate the probability that a person who comes into the store will buy nothing.

Of the 56 people whom came into the store, 23 bought something. This means that 56-23 = 33 of them did not buy anything. So:

$D = 33, T = 56$

$P = \frac{D}{T} = \frac{33}{56} = 0.5893$

There is a 58.93% probability that a person who comes into the store will buy nothing.

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