All categories

2 years ago

Y=2x^2-4x-3 Solve quadratic questions

1 answer:

6 0

By applying the quadratic formula we conclude that the roots of the quadratic equation y = 2 · x² - 4 · x - 3 are 1 + 0.5√10 and 1 - 0.5√10, respectively.

<h3>How to find the roots of a quadratic equation </h3>

All quadratic equations of the form a · x² + b · x + c = 0 have two roots that can be found by using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^{2}-4\cdot a \cdot c}}{2\cdot a}$ (1)

If we know that a = 2, b = - 4 and c = - 3, then the roots of the quadratic equations are:

$x = \frac{4 \pm \sqrt{(-4)^{2}-4\cdot (2) \cdot (- 3)}}{2\cdot (2)}$

$x = 1 \pm \frac{\sqrt{40}}{4}$

$x = 1 \pm \frac{\sqrt{10}}{2}$

By applying the quadratic formula we conclude that the roots of the quadratic equation y = 2 · x² - 4 · x - 3 are 1 + 0.5√10 and 1 - 0.5√10, respectively.

To learn more on quadratic equations: brainly.com/question/2263981

#SPJ1

You might be interested in

2 3/5 * 3 1/3 Just answer please

Ira Lisetskai [31]

Answer: 8.66666666667

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

The daily low temperatures last week are shown below. What is the mean

kirill115 [55]

Answer:

i do not see the tempatures but add them all up and divide by however many tempatures there are

Step-by-step explanation:

5 0

3 years ago

How to determine if the line is parallel , perpendicular, or neither

4vir4ik [10]

It is parallel if the lines are in straight line next to each other, but don't cross.
It is perpendicular if these lines intersect to form right angles.
Neither if they don' t follow these rules

5 0

3 years ago

Use lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f(x,y = xyz; x^

Snezhnost [94]

I'm assuming the constraint involves some plus signs that aren't appearing for some reason, so that you're finding the extrema subject to $x^2+2y^2+3z^2=96$ .

Set $f(x,y,z)=xyz$ and $g(x,y,z)=x^2+2y^2+3z^2-96$ , so that the Lagrangian is

$L(x,y,z,\lambda)=xyz+\lambda(x^2+2y^2+3z^2-96)$

Take the partial derivatives and set them equal to zero.

$\begin{cases}L_x=yz+2\lambda x=0\\L_y=xz+4\lambda y=0\\L_z=xy+6\lambda z=0\\L_\lambda=x^2+2y^2+3z^2-96=0\end{cases}$

One way to find the possible critical points is to multiply the first three equations by the variable that is missing in the first term and dividing by 2. This gives

$\begin{cases}\dfrac{xyz}2+\lambda x^2=0\\\\\dfrac{xyz}2+2\lambda y^2=0\\\\\dfrac{xyz}2+3\lambda z^2=0\\\\x^2+2y^2+3y^2=96\end{cases}$

So by adding the first three equations together, you end up with

$\dfrac32xyz+\lambda(x^2+2y^2+3z^2)=0$

and the fourth equation allows you to write

$\dfrac32xyz+96\lambda=0\implies \dfrac{xyz}2=-32\lambda$

Now, substituting this into the first three equations in the most recent system yields

$\begin{cases}-32\lambda+\lambda x^2=0\\-32\lambda+2\lambda y^2=0\\-32\lambda+3\lambda z^2=0\end{cases}\implies\begin{cases}x=\pm4\sqrt2\\y=\pm4\\z=\pm4\sqrt{\dfrac23}\end{cases}$

So we found a grand total of 8 possible critical points. Evaluating $f(x,y,z)=xyz$ at each of these points, you find that $f(x,y,z)$ attains a maximum value of $\dfrac{128}{\sqrt3}$ whenever exactly none or two of the critical points' coordinates are negative (four cases of this), and a minimum value of $-\dfrac{128}{\sqrt3}$ whenever exactly one or all of the critical points' coordinates are negative.

6 0

3 years ago

Identify the horizontal asymptote of f(x) =x2+5x-3/4x-1

Pachacha [2.7K]

since the numerator is x² + 5x - 3, and therefore has a degree of 2, whilst the denominator, 4x¹ - 1, has a degree of 1, therefore, there's no horizontal asymptote.

recall, we only get a horizontal asymptote if the denominator's expression degree is equals or greater than that of the numerator's.

5 0

4 years ago