All categories

3 years ago

Use the parabola tool to graph the quadratic function f(x) =-5x2 -2

Mathematics

1 answer:

enot [183]3 years ago

3 0

Your first point would be on (0, 2), the vertex. Then, put a point on (-2, 22). Hope that helps!

You might be interested in

Definition: X and Y coordinates???????????????????

kvv77 [185]

Answer:

The x coordinate is a given number of pixels along the horizontal axis of a display starting from the pixel (pixel 0) on the extreme left of the screen. The y coordinate is a given number of pixels along the vertical axis of a display starting from the pixel (pixel 0) at the top of the screen.

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Find the measure of ∠ADB ∠ A D B if ∠ADC = 82°

sveta [45]

Assuming CDB is a straight line then angle ADB = 180-82=98

8 0

3 years ago

Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?

cluponka [151]

Under the given transformation, the Jacobian and its determinant are

$\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2$

so that

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv$

where $D'$ is the region $D$ transformed into the $u$ - $v$ plane. The remaining integral is the twice the area of $D'$ .

Now, the integral over $D$ is

$\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y$

but through the given transformation, the boundary of $D'$ is the set of equations,

$\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}$

We require that $u>0$ , and the last equation tells us that we would also need $v>0$ . This means $1\le v\le\sqrt2$ and $0$ , so that the integral over $D'$ is

$\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6$

4 0

3 years ago

Using the slope formula, determine the slope of the two points.

Lelechka [254]

The slope formula is is rise over run or y2-y1/x2-x1

If you plug in those numbers you will get

(-1)-(-4)
———
1-(-2)

Simplify it and you get

3
— or 1
3

So the slope is 1

5 0

3 years ago

F(x)= 6+ (10/x) what are the inflection points?

mojhsa [17]

Answer:

An inflection point is a point on the graph of a function at which the concavity changes. Points of inflection can occur where the second derivative is zero.

Step-by-step explanation:

7 0

3 years ago