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2 years ago

Analyze the graph of the cube root function shown

Mathematics

1 answer:

Greeley [361]2 years ago

7 0

The values of a, h, and k in the general equation is h=1, k=4 and a=2

<h3>What is parent function?</h3>

The function which can be transformed into different ways is the parent function.

On a coordinate plane, a cubic function approaches the x-axis in quadrant 2, crosses the y-axis at (0, 2), has an inflection point at (1, 4), and the goes through (2, 6).

The given function is

y =a³ √(x -h) + k

Put the coordinate (0,2) in the equation, we get

2 = a³ √(0 -h) + k..................(1)

Plug (1, 4), we have

4 = a³ √(1 -h) + k..................(2)

Plug (2, 6), we have

6 = a³ √(2 -h) + k..................(3)

On solving these three equations, we have

h=1

k=4

a=2

Thus, value of h is 1, k is 4 and a is 2.

Learn more about parent function

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3 years ago

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-6 1/8

Step-by-step explanation:

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as a membership fee, a health club charges a one time amount of $40 and charges $25 each month. the total fee after m months is

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3 years ago

Complete the equation of the line through (-6,5)(−6,5)left parenthesis, minus, 6, comma, 5, right parenthesis and (-3,-3)(−3,−3)

sladkih [1.3K]

Answer:

The equation of the line passing through the given points is:

$y=-\frac{8}{3}x-11$

Step-by-step explanation:

Given:

The two points are: $(x_1,y_1)=(-6,5)\textrm{ and }(x_2,y_2)=(-3,-3)$

The equation of a line when two points are given is:

$y-y_1=\left (\frac{y_2-y_1}{x_2-x_1} \right )(x-x_1)$

Plug in all the values and simplify.

$y-5=\left (\frac{-3-5}{-3-(-6)} \right )(x-(-6))\\y-5=\left (\frac{-8}{-3+6} \right )(x+6)\\y-5=\left (\frac{-8}{3} \right )(x+6)\\y-5=\frac{-8}{3}x-16\\y=\frac{-8}{3}x-16+5\\\\y=-\frac{8}{3}x-11$

Therefore, the equation of the line passing through the above points is

$y=-\frac{8}{3}x-11$

7 0

3 years ago

Find the mass of the triangular region with vertices (0, 0), (3, 0), and (0, 1), with density function ρ(x,y)=x2+y2.

ololo11 [35]

Since density is the ratio of mass to (in this case) area, we can find the mass of the triangular region $\mathcal T$ by computing the double integral of the density function over $\mathcal T$ :

$\mathrm{mass}=\displaystyle\iint_{\mathcal T}\rho(x,y)\,\mathrm dx\,\mathrm dy$

The boundary of $\mathcal T$ is determined by a set of lines in the $x,y$ plane. One way to describe the region $\mathcal T$ is by the set of points,

$\mathcal T=\left\{(x,y)\mid0\le x\le 3\,\land\,0\le y\le1-\dfrac x3\right\}$

So the mass is

$\mathrm{mass}=\displaystyle\int_{x=0}^{x=3}\int_{y=0}^{y=1-x/3}(x^2+y^2)\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_{x=0}^{x=3}\left(x^2y+\frac{y^3}3\right)\bigg|_{y=0}^{y=1-x/3}\,\mathrm dx$

$=\displaystyle\int_{x=0}^{x=3}\left(x^2\left(1-\frac x3\right)+\frac{\left(1-\frac x3\right)^3}3\right)\,\mathrm dx$

$=\displaystyle\frac1{81}\int_0^3(27-27x+90x^2-28x^3)\,\mathrm dx=\frac52$

6 0

3 years ago