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Vadim26 [7]
3 years ago
14

write an equation for a function that has a graph with the shape of y=x^2, but upsidedown and shifted right 6 units and up 4 uni

ts. ​
Mathematics
1 answer:
irina1246 [14]3 years ago
5 0

Answer:

y - 4 = -(x - h)^2

Step-by-step explanation:

We begin with the stock function y = x^2.  Its graph is a parabola with vertex at (0, 0) and opening up.  The more general vertex form of the graph is

y - k = a(x - h)^2, where the point (h, k) is the vertex and 'a' determines the direction of opening as well as by how much the parabola is stretched vertically.

In this problem the vertex is (6, 4); that is, the vertex is shifted to the right by 6 units and upward from there by 4 units.  The coefficient 'a' is -1, indicating that the graph opens downward.

So we have the equation y - 4 = -(x - h)^2.  Without more information we cannot tell whether the graph is stretched vertically or not.

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Answer:

ax² + bx + c

Step-by-step explanation:

The form of a quadratic equation that is easy to use when finding the maximum or minimum value of the function is ax² + bx + c.

Suppose a quadratic function:

f(x) = 2x² - 8x + 9

Use ( -b/2a ,  f(-b/2a) ).

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-(-8)/2(2)

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f(2) = 2(2)² - 8(2) + 9

f(2) = 2(4) - 8(2) + 9

f(2) = 8 - 16 + 9

f(2) = 1

The minimum value of this quadratic function is (2, 1).

It represents a minimum value because a > 0.

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How do you find the Constant of variation
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If y
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varies directly as x
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and y=15
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when x=24
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k=yx=1524=58y=58x k=
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y=
5
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Hello please help i’ll give brainliest
tamaranim1 [39]

Answer: The answer is D

Step-by-step explanation: It is the only option that makes sense to me.

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<em>Is that clear?</em>

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