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zvonat [6]
4 years ago
10

Help me please! Explain how you can use the graph of the quadratic function f(x) to solve the equation f(x)=0

Mathematics
2 answers:
bulgar [2K]4 years ago
6 0
Ok so:  the graph of a function intersects the x-axis will be the roots of f(x) = 0. Because the coordinates of the point where it intersects the xaxis will be (x, 0). y will be 0. What you need to do is add 0 wherever x is for eample: <span>when multiplying 3 and 0 you get 0 then subtract 2 and y would equal negative 2. I hope this helps you</span>
DochEvi [55]4 years ago
5 0

Sample Response: The zeros of a quadratic function are the solutions of the related quadratic equation. If the graph of a quadratic function crosses the x-axis, there will be two real number solutions. If the graph of a quadratic function just touches the x-axis, there will be one unique real number, or double root, solution. If the graph of a quadratic function does not cross the x-axis, there will be no real number solution.  Hope it helps! Thank you!

-Charlie



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43.96 over 14

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Simplify the given polynomial and use it to complete the statement (8x^2 - 5x -7) - (9x^2 - 3x -5) + (x+7)(x + 1)
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Answer:

The answer to your question is:    6x + 5

Step-by-step explanation:

                          (8x² - 5x -7) - (9x² - 3x -5) + (x+7)(x + 1)

                          8x² - 5x - 7 - 9x² + 3x + 5 + x² + 8x + 7

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8 0
3 years ago
1. Write the polynomial function that models the given situation.A rectangle has a length of 12 units and a width of 11 units. S
kolbaska11 [484]

Answer:

1.  (12 - 2x)(11 - 2x)x

2. 4(11 - 2x)²(x + 1)

3. π(x³ + 15x² + 63x + 81)

Step-by-step explanation:

1. Write the polynomial function that models the given situation.

A rectangle has a length of 12 units and a width of 11 units. Squares of x by x units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume V of the box as a polynomial function in terms of x.

Since the length of the rectangle is 12 units and its width 11 units and squares of x by x units are cut from its corners, it implies that a length x is cut from each side. So, the length of the open box is L = 12 - 2x and its width is w = 11 - 2x.

Since the cut corners of the rectangle are folded, the side x which is cut represents the height of the open box, h. so, h = x

So, the volume of the open box V = LWh = (12 - 2x)(11 - 2x)x

2. Write the polynomial function that models the given situation. A square has sides of 24 units. Squares x + 1 by x + 1 units are cut out of each corner, and then the sides are folded up to create an open box. Express the volume V of the box as a function in terms of x.

Since the square has sides of 24 units and squares of x + 1 by x + 1 units are cut from its corners, it implies that a length x + 1 is cut from each corner and the length 2(x + 1) is cut from each side. So, the length of side open box is L = 24 - 2(x + 1) = 24 - 2x - 2 = 24 - 2 - 2x = 22 - 2x = 2(11  - x)

Since the cut corners of the square are folded, the side x + 1 which is cut represents the height of the open box, h. so, h = x + 1

Since the area of the base of the pen box is a square, its area is L² = [2(11 - 2x)]²

So, the volume of the open box V = L²h = [2(11 - 2x)]²(x + 1) = 4(11 - 2x)²(x + 1)

3. Write the polynomial function that models the given situation. A cylinder has a radius of x + 6 units and a height 3 units more than the radius. Express the volume V of the cylinder as a polynomial function in terms of x.

The volume of a cylinder is V = πr²h where r = radius and h = height of cylinder.

Given that r = x + 6 and h is 3 units more than r, h = r + 3 = x + 6 + 3 = x + 9

So, V = πr²h

V = π(x + 3)²(x + 9)

V = π(x² + 6x + 9)(x + 9)

V = π(x³ + 6x² + 9x + 9x² + 54x + 81)

V = π(x³ + 15x² + 63x + 81)

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