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Fudgin [204]
3 years ago
12

A jumping spider's movement is modeled by a parabola. The spider makes a single jump from the origin and reaches a maximum heigh

t of 10 mm halfway across a horizontal
distance of 80 mm.
Part A: Write the equation of the parabola in standard form that models the spider's jump. Show your work. (4 points)
Part B: Identify the focus, directrix, and axis of symmetry of the parabola. (6 points)
Mathematics
1 answer:
Stella [2.4K]3 years ago
6 0

A parabola is a mirror-symmetrical U-shape.

  • The equation of the parabola is \mathbf{y = -\frac{1}{640}(x - 80)^2 + 10}
  • The focus is \mathbf{Focus = (80, -1760)}
  • The directrix is \mathbf{y = \frac{1}{640}}
  • The axis of the symmetry of parabola is: \mathbf{x = 80}

From the question, we have:

\mathbf{Vertex: (h,k) = (80,10)}

\mathbf{Origin: (x,y) = (0,0)}

The equation of a parabola is:

\mathbf{y = a(x - h)^2 + k}

Substitute the values of origin and vertex in \mathbf{y = a(x - h)^2 + k}

\mathbf{0 = a(0 - 80)^2 + 10}

\mathbf{0 = a(- 80)^2 + 10}

\mathbf{0 = 6400a + 10}

Collect like terms

\mathbf{6400a =- 10}

Solve for a

\mathbf{a =- \frac{1}{640}}

Substitute the values of a and the vertex in \mathbf{y = a(x - h)^2 + k}

\mathbf{y = -\frac{1}{640}(x - 80)^2 + 10}

The focus of a parabola is:

\mathbf{Focus = (h, \frac{k+1}{4a})}

Substitute the values of a and the vertex in \mathbf{Focus = (h, \frac{k+1}{4a})}

\mathbf{Focus = (80, \frac{10+1}{4 \times -\frac{1}{640}})}

\mathbf{Focus = (80, -\frac{11}{\frac{1}{160}})}

\mathbf{Focus = (80, -11\times 160)}

\mathbf{Focus = (80, -1760)}

The equation of the directrix is:

\mathbf{y = -a}

So, we have:

\mathbf{y = \frac{1}{640}} ----- the directrix

The axis of symmetry is:

\mathbf{x = -\frac{b}{2a}}

We have:

\mathbf{y = -\frac{1}{640}(x - 80)^2 + 10}

Expand

\mathbf{y = -\frac{1}{640}(x^2 -160x + 6400) +10}

Expand

\mathbf{y = -\frac{1}{640}x^2 +\frac{1}{4}x - 10 +10}

\mathbf{y = -\frac{1}{640}x^2 +\frac{1}{4}x }

A quadratic function is represented as:

\mathbf{y = ax^2 + bx + c}

So, we have:

\mathbf{a = -\frac{1}{640}}

\mathbf{b = \frac{1}{4}}

Recall that:

\mathbf{x = -\frac{b}{2a}}

So, we have:

\mathbf{x = -\frac{1/4}{2 \times -1/640}}

\mathbf{x = \frac{1/4}{1/320}}

This gives

\mathbf{x = \frac{320}{4}}

\mathbf{x = 80}

Hence, the axis of the symmetry of parabola is: \mathbf{x = 80}

Read more about parabola at:

brainly.com/question/21685473

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The Students' Conjectures Emily and Zach have two different polynomials to multiply: Polynomial product A: (4x2 – 4x)(x2 – 4) Po
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Complete Question:

Emily and Zach have two different polynomials to multiply: Polynomial product A: (4x2 – 4x)(x2 – 4) Polynomial product B: (x2 + x – 2)(4x2 – 8x) They are trying to determine if the products of the two polynomials are the same. But they disagree about how to solve this problem.

Answer:

A = B

Step-by-step explanation:

<em>See comment for complete question</em>

Given

A: (4x^2 - 4x)(x^2 - 4)

B: (x^2 + x - 2)(4x^2 - 8x)

Required

Determine how they can show if the products are the same or not

To do this, we simply factorize each polynomial

For, Polynomial A: We have:

A: (4x^2 - 4x)(x^2 - 4)

Factor out 4x

A: 4x(x - 1)(x^2 - 4)

Apply difference of two squares on x^2 - 4

A: 4x(x - 1)(x - 2)(x+2)

For, Polynomial B: We have:

B: (x^2 + x - 2)(4x^2 - 8x)

Expand x^2 + x - 2

B:(x^2 + 2x - x - 2)(4x^2- 8x)

Factorize:

B:(x(x + 2) -1(x + 2))(4x^2- 8x)

Factor out x + 2

B:(x -1) (x + 2)(4x^2- 8x)

Factor out 4x

B:(x -1) (x + 2)4x(x- 2)

Rearrange

B: 4x(x - 1)(x - 2)(x+2)

The simplified expressions are:

A: 4x(x - 1)(x - 2)(x+2) and  

B: 4x(x - 1)(x - 2)(x+2)

Hence, both polynomials are equal

A = B

3 0
3 years ago
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