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3 years ago

The parabola y=x² is shifted to the right 8 units. What is the equation to the new parabola

Mathematics

1 answer:

JulijaS [17]3 years ago

4 0

The new equation is

<em>Y = (x-8)² </em>.

The other way to write it is like this: (you might not recognize it in this form)

Y = x² - 16x + 64

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What are the features of the parabola modeled by the equation f(x) = x2 - 114x ?

notsponge [240]

Answer:

Answers are in bold type

Step-by-step explanation:

f(x) = $x^{2} -144x$

The parabola opens up, so has a minimum at the vertex.

Let (h, k) be the vertex

h = -b/2a = - (-144)/2(1) = 57

k = 57^2 - 144(57) = 3249 - 6498 = -3249

Therefore, the vertex is (57, -3249)

The minimum value is -3249

The domain is the set of real numbers.

The range = {y | y ≥ -3249}

The function decreases when -∞ < x < 57 and increases when 57 > x > ∞

The x - intercepts: $x^{2} -144x$ = 0

x(x - 114x) = 0

x = 0 or x = 114

x-intercepts are (0, 0) and (0, 114)

When x = 0, then we get the y-intercept. So, 0^2 - 114(0) = 0

y-intercept is (0, 0)

3 0

3 years ago

Evaluate 2x2 - 1 when x = 3.

tigry1 [53]

X should equal 11
2(3)2-1
6(2)-1
12-1
11

6 0

3 years ago

Read 2 more answers

This is my last question

mariarad [96]

Answer: B

Step-by-step explanation:because my teacher said she having trouble

5 0

3 years ago

F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa

DerKrebs [107]

a. Parameterize $C$ by

$\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath$

with $0\le t\le1$ .

b/c. The line integral of $\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath$ over $C$ is

$\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt$

$=\displaystyle10\int_0^1\mathrm dt=\boxed{10}$

d. Notice that we can write the line integral as

$\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)$

By Green's theorem, the line integral is equivalent to

$\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy$

where $D$ is the triangle bounded by $C$ , and this integral is simply twice the area of $D$ . $D$ is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

4 0

3 years ago

What is the answer of this question ?

fgiga [73]

Answer:

the answer is B

Step-by-step explanation:

hoped I helped:)

6 0

2 years ago