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).

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?

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

i dont know sorry

step-by-step explanation: