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2 years ago

Find the x- and y-intercept of the parabola y = x² + 4x + 2

Mathematics

1 answer:

Veronika [31]2 years ago

4 0

Answer:

x-intercept = (- 2 + √2, 0) (- 2 - √2, 0)

y-intercept = (2, 0)

Step-by-step explanation:

<u>Given expression</u>

y = x² + 4x + 2

<u>Find the x-intercept</u>

<em>Definition of x-intercept is the point that intersects with the x-axis, which means the y value is 0</em>

(0) = x² + 4x + 2

(x + 2 - √2) (x + 2 + √2) = 0

x = -2 ± √2

Therefore, the x-intercepts are $\Large\boxed{(-2+\sqrt{2},~0) }$ $\Large\boxed{(-2-\sqrt{2},~0) }$

<u>Find the y-intercept</u>

<em>Definition of x-intercept is the point that intersects with the y-axis, which means the x value is 0</em>

y = (0)² + 4(0) + 2

y = 0 + 0 + 2

y = 2

Therefore, the y-intercept is $\Large\boxed{(0,~2)}$

Hope this helps!! :)

Please let me know if you have any questions

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goblinko [34]

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If ABC is congruent to DEF
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3 years ago

Read 2 more answers

Factor 2x4 - 20x2 - 78.

Airida [17]

Answer:

2x⁴ - 20x² - 78

To factor the expression look for the LCM of the numbers

LCM of the numbers is 2

Factorize that one out

That's

2( x⁴ - 10x² - 39)

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3 years ago

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Which expression is equivalent to (16 x Superscript 8 Baseline y Superscript negative 12 Baseline) Superscript one-half?.

loris [4]

To solve the problem we must know the Basic Rules of Exponentiation.

<h2>Basic Rules of Exponentiation</h2>

$x^ax^b = x^{(a+b)}$
$\dfrac{x^a}{x^b} = x^{(a-b)}$
$(a^a)^b =x^{(a\times b)}$
$(xy)^a = x^ay^a$

$x^{\frac{3}{4}} = \sqrt[4]{x^3}= (\sqrt[3]{x})^4$

The solution of the expression is $\dfrac{4x^4}{y^6}$ .

<h2>Explanation</h2>

Given to us

$(16x^8y^{12})^{\frac{1}{2}}$

Solution

We know that 16 can be reduced to $2^4$ ,

$=(2^4x^8y^{12})^{\frac{1}{2}}$

Using identity $(xy)^a = x^ay^a$ ,

$=(2^4)^{\frac{1}{2}}(x^8)^{\frac{1}{2}}(y^{12})^{\frac{1}{2}}$

Using identity $(a^a)^b =x^{(a\times b)}$ ,

$=(2^{4\times \frac{1}{2}})\ (x^{8\times\frac{1}{2}})\ (y^{12\times{\frac{1}{2}}})$

Solving further

$=2^2x^4y^{-6}$

Using identity $\dfrac{x^a}{x^b} = x^{(a-b)}$ ,

$=\dfrac{2^2x^4}{y^6}$

$=\dfrac{4x^4}{y^6}$

Hence, the solution of the expression is $\dfrac{4x^4}{y^6}$ .

Learn more about Exponentiation:

brainly.com/question/2193820

8 0

2 years ago

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

$\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]$

The left and right endpoints of the $i$ -th subinterval, respectively, are

$\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8$

$r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8$

for $1\le i\le4$ , and the respective midpoints are

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8$

Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

$T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}$

Midpoint rule

We approximate the area for each subinterval by

$M_i=f(m_i)(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}$

Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial $p_i(x)$ , where

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It so happens that the integral of $p_i(x)$ reduces nicely to the form you're probably more familiar with,

$S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))$

Then the integral is approximately

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}$

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

3 0

3 years ago

9+8{7•6-5[4+(3-2•1)]}=?

34kurt

Answer:

145

Step-by-step explanation:

Ⓗⓘ ⓣⓗⓔⓡⓔ

9+8{7*6-5[4+(3-2*1)}]}

9+8{7*6-5[4+1]}

9+8{7*6-25}

9+8{42-25}

9+136

145

(っ◔◡◔)っ ♥ Hope this helped! Have a great day! :) ♥

Please, please give brainliest, it would be greatly appreciated, I only a few more before I advance, thanks!

5 0

3 years ago