Given ΔBAE is isosceles with base angles ∠BEA and ∠EBA and that ∠AEC = 115°, is ΔEBC ~ ΔAEC? If so, by what criterion?
1 answer:
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Answer:
(1, 3)
Step-by-step explanation:
You are given the h coordinate of the vertex as 1, but in order to find the k coordinate, you have to complete the square on the parabola. The first few steps are as follows. Set the parabola equal to 0 so you can solve for the vertex. Separate the x terms from the constant by moving the constant to the other side of the equals sign. The coefficient HAS to be a +1 (ours is a -2 so we have to factor it out). Let's start there. The first 2 steps result in this polynomial:
. Now we factor out the -2:
. Now we complete the square. This process is to take half the linear term, square it, and add it to both sides. Our linear term is 2x. Half of 2 is 1, and 1 squared is 1. We add 1 into the set of parenthesis. But we actually added into the parenthesis is +1(-2). The -2 out front is a multiplier and we cannot ignore it. Adding in to both sides looks like this:
. Simplifying gives us this:
On the left we have created a perfect square binomial which reflects the h coordinate of the vertex. Stating this binomial and moving the -3 over by addition and setting the polynomial equal to y:
From this form,
you can determine the coordinates of the vertex to be (1, 3)