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ollegr [7]
3 years ago
5

The graph of y = f(x) has the line x = 5 as an axis of symmetry. The graph also passes through the point (8,-7). Find another po

int that must lie on the graph of y = f(x).
Mathematics
1 answer:
qaws [65]3 years ago
3 0

You are missing an entire paragraph of information that is needed to solve what you've actually posted. Luckily for you, I answered this question once before.

Here is the ENTIRE QUESTION:

The graph of y = f(x) is a parabola whose vertex is at (1, -2). The graph of y = f(x - 3) is also a parabola. Where is the vertex of the graph of y = f(x-3)? Write your answer as an ordered pair.

Solution:

The "f( x - 3)" shifts the the graph of f(x) to the right by 3 units.

So...the vertex of f(x - 3) is ( 4, - 2).

2. The graph of y = f(x) has the line x = 5 as an axis of symmetry. The graph also passes through the point (8,-7). Find another point that must lie on the graph of y = f(x).

Here is the solution you are looking for.

If the graph has an axis of symmetry of x = 5 and (8, -7) is on the graph, we simply add 3 to 5 to get (8,-7). Afterwards, we must subtract 3 from 5 to get : (5 - 3, -7) = (2, -7) is also on the graph.

Notice that the y-coordinate is unaffected. Cool, right?

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Sidana [21]
If we have 

y = 6x + 1 and x - y = 11, we can take advantage of  y - y = 0 as follows:
                                         Subtract x from both sides of the 2nd equation:
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--------------
0 = 5x + 12.  Solving for x, x = -12/5.     Using the equation above, find y:
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                                                                   = 11 - (-12/5)
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3 0
4 years ago
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3. The side can never be equal to radius of circle because the side of the square will be chord for the circle and in a circle chord can never be equal to its radius


7 0
3 years ago
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