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3 years ago

Find the vertex of the parabola given by y=2x^2-4x-1

1 answer:

3 0

Given the quadratic y = 2x^2-4x-1, I identify A = 2, B= -4 and C = -1. The axis of symmety: x = -b/2a --> 4/2(2) = 4/4 = 1. Therefore, the x-coordinate of the vertex is 1. To find the y-coordinate, I plug in 1 for x into the quadratic. Hence, y = 2(1)^2-4(1)-1 ---> -3. The coordinates of the vertex are (1,-3).

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Suppose a rectangular pasture is to be constructed using 1 2 linear mile of fencing. The pasture will have one divider parallel

timama [110]

Answer:

$\displaystyle A=\frac{1}{192}$

Step-by-step explanation:

<u>Maximization With Derivatives</u>

Given a function of one variable A(x), we can find the maximum or minimum value of A by using the derivatives criterion. If A'(x)=0, then A has a probable maximum or minimum value.

We need to find a function for the area of the pasture. Let's assume the dimensions of the pasture are x and y, and one divider goes parallel to the sides named y, and two dividers go parallel to x.

The two divisions parallel to x have lengths y, thus the fencing will take 4x. The three dividers parallel to y have lengths x, thus the fencing will take 3y.

The amount of fence needed to enclose the external and the internal divisions is

$P=4x+3y$

We know the total fencing is 1/2 miles long, thus

$\displaystyle 4x+3y=\frac{1}{2}$

Solving for x

$\displaystyle x=\frac{\frac{1}{2}-3y}{4}$

The total area of the pasture is

$A=x.y$

Substituting x

$\displaystyle A=\frac{\frac{1}{2}-3y}{4}.y$

$\displaystyle A=\frac{\frac{1}{2}y-3y^2}{4}$

Differentiating with respect to y

$\displaystyle A'=\frac{\frac{1}{2}-6y}{4}$

Equate to 0

$\displaystyle \frac{\frac{1}{2}-6y}{4}=0$

Solving for y

$\displaystyle y=\frac{1}{12}$

And also

$\displaystyle x=\frac{\frac{1}{2}-3\cdot \frac{1}{12}}{4}=\frac{1}{16}$

Compute the second derivative

$\displaystyle A''=-\frac{3}{2}.$

Since it's always negative, the point is a maximum

Thus, the maximum area is

$\displaystyle A=\frac{1}{12}\cdot \frac{1}{16}=\frac{1}{192}$

6 0

3 years ago

2(g+5)=-7-4g+g include instructions plz!

Elza [17]

First distribute the 2, then add 7 to that side, so you have 2g+17=-4g+g , subtract 2g and it should be easy to find from there lol

4 0

3 years ago

If m∠SRQ = 6x + 2 and m∠QRP = 2x + 10, find the value of x . *

Neko [114]

Answer:

for m∠SRQ x = 3

for m∠QRP x = 5

Step-by-step explanation:

1. 6x +2. 6 is being multiplied by a number so to find that number we do teh opposite

6x/2 = 3

drop the x then u have x = 3

2. 2x+10. 2 is being muliplied by a number so to find that number we do teh opposite

10/2x = 5

drop the x then u have x = 5

7 0

3 years ago

Two experiments are defined below. An event is defined for each of the experiments. Experiment I: Corrine rolls a standard six-s

-BARSIC- [3]

Answer: The correct answer is option C: Both events are equally likely to occur

Step-by-step explanation: For the first experiment, Corrine has a six-sided die, which means there is a total of six possible outcomes altogether. In her experiment, Corrine rolls a number greater than three. The number of events that satisfies this condition in her experiment are the numbers four, five and six (that is, 3 events). Hence the probability can be calculated as follows;

P(>3) = Number of required outcomes/Number of possible outcomes

P(>3) = 3/6

P(>3) = 1/2 or 0.5

Therefore the probability of rolling a number greater than three is 0.5 or 50%.

For the second experiment, Pablo notes heads on the first flip of a coin and then tails on the second flip. for a coin there are two outcomes in total, so the probability of the coin landing on a head is equal to the probability of the coin landing on a tail. Hence the probability can be calculated as follows;

P(Head) = Number of required outcomes/Number of all possible outcomes

P(Head) = 1/2

P(Head) = 0.5

Therefore the probability of landing on a head is 0.5 or 50%. (Note that the probability of landing on a tail is equally 0.5 or 50%)

From these results we can conclude that in both experiments , both events are equally likely to occur.

3 0

3 years ago

PLEASE HELP!! Compare & Contrast how Completing the Square is used to Convert a quadratic function to vertex form with how i

Likurg_2 [28]

Answer:

Step-by-step explanation:

Given a general quadratic formula given as ax²bx+c = 0

To generate the general formula to solve the quadratic equation, we can use the completing the square method as shown;

Step 1:

Bringing c to the other side

ax²+bx = -c

Dividing through by coefficient of x² which is 'a' will give:

x²+(b/a)x = -c/a

- Completing the square at the left hand side of the equation by adding the square of half the coefficient x i.e (b/2a)² and adding it to both sides of the equation we have:

x²+(b/a)x+(b/2a)² = -c/a+(b/2a)²

(x+b/2a)² = -c/a+(b/2a)²

(x+b/2a)² = -c/a + b²/4a²

- Taking the square root of both sides

√(x+b/2a)² = ±√-c/a + b²/√4a²

x+b/2a = ±√(-4ac+b²)/√4a²

x+b/2a =±√b²-4ac/2a

- Taking b/2a to the other side

x = -b/2a±√√b²-4ac/2a

Taking the LCM:

x = {-b±√b²-4ac}/2a

This gives the vertex form with how it is used to Solve a quadratic equation.

7 0

3 years ago