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horrorfan [7]
2 years ago
13

The graph represents a cubic function. Write the function.

Mathematics
2 answers:
Norma-Jean [14]2 years ago
7 0

Zeros are

  • (0,0)
  • (-3,0)
  • (-4,0)

Write in factor form and solve

  • (x-0)(x-(-3))(x-(-4))=0
  • x(x+3)(x+4)=0
  • x[x(x+4)+3(x+4)]=0
  • x(x²+4x+3x+12)=0
  • x(x²+7x+12)=0
  • x³+7x²+12x=0

Points given (-2,16)

Solve for n

  • n(x³+7x²+12x)=y

On solving we get

  • n=-4

So the function is

  • y=-4(x³+7x²+12x)
  • y=-4x³-28x²-48x
TiliK225 [7]2 years ago
4 0

Answer:

-4x³ - 28x²- 48x

Explanation:

  • x-intercept: (-4, 0), (-3, 0), (0, 0)

→ so equation: a(x+4)(x+3)(x+0)

  • point given: (-2, 16)

→ substitute these points to find <u>a</u>

  • a(-2+4)(-2+3)(-2) = 16
  • a = -4

so the graphed cubic equation:

  • -4(x+4)(x+3)(x+0)
  • -4x³ -28x²-48x
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fomenos

Answer:

z=1.96

Step-by-step explanation:

Using normal distribution table or technology, 97.5% corresponds to z=1.959964, generally denoted z=1.96, or 1.96 standard deviations above the mean.

(above value obtained from R)

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Part A:

The average rate of change refers to a function's slope. Thus, we are going to need to use the slope formula, which is:

m = \dfrac{y_2 - y_1}{x_2 - x_1}

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You can see that we are given the x-values for our interval, but we are not given the y-values, which means that we will need to find them ourselves. Remember that the y-values of functions refers to the outputs of the function, so to find the y-values simply use your given x-value in the function and observe the result:

h(0) = 3(5)^0 = 3 \cdot 1 = 3

h(1) = 3(5)^1 = 3 \cdot 5 = 15

h(2) = 3(5)^2 = 3 \cdot 25 = 75

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Now, let's find the slopes for each of the sections of the function:

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m = \dfrac{15 - 3}{1 - 0} = \boxed{12}

<u>Section B</u>

m = \dfrac{375 - 75}{3 - 2} = \boxed{300}


Part B:

In this case, we can find how many times greater the rate of change in Section B is by dividing the slopes together.

\dfrac{m_B}{m_A} = \dfrac{300}{12} = 25


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Answer:

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we know that

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y-y1=m(x-x1)

In this problem we have

point\ (4,\frac{1}{3})

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substitute the given values

y-\frac{1}{3}=\frac{3}{4}(x-4)

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