Answer:
y = -7(x +1)² +35
Step-by-step explanation:
The given quadratic equation has two parameters you are asked to find. When you fill in the given point values, the result is two linear equations in the two unknown parameters. You can solve these equations in any of the usual ways.
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<h3>equation for point (1, 7)</h3>y = a(x +1)² +k
7 = a(1 +1)² +k
7 = 4a +k
<h3>equation for point (2, -28)</h3>-28 = a(2 +1)² +k
-28 = 9a +k
<h3>solution</h3>The first equation can be subtracted from the second equation to eliminate the k variable. (This is solving by <em>elimination</em>.)
(-28) -(7) = (9a +k) -(4a +k)
-35 = 5a . . . . . . . . simplify
-7 = a . . . . . . . . divide by 5
Using the first equation, we can find k.
7 -4a = k . . . . . subtract 4a
7 -4(-7) = k . . . . substitute for 'a'
35 = k
The parameter values are a = -7, and k = 35. The quadratic equation is ...
y = -7(x +1)² +35
Let P be a point outside the circle such that triangle LMP has legs coincident with chords MW and LK (i.e. M, W, and P are colinear, and L, K, and P are colinear). By the intersecting secants theorem,
The angles in any triangle add to 180 degrees in measure, and and
, so that