Write three different equations that have x=5 as a solution
2 answers:
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Answer:
Step-by-step explanation:
A sequence is defined by the recursive formula
If then
for
for
for
for
Answer:
Step-by-step explanation:
In order to determine the information you're being asked for, you need to complete the square on that quadratic. The first step is to move the constant over to the other side of the equals sign:
Here would be the step where, if the leading coefficient isn't a 1, you'd factor it out. But ours is a 1, so we're good there. Now take half the linear term (the term with the single x on it), square it, and add it to both sides. Our linear term is a -2. Half of -2 is -1, and -1 squared is +1. We add +1 to both sides giving us this:
Now we'll clean it up a bit. The right side becomes a 4, and the left side is written as its perfect square binomial, which is the whole reason we did this. That binomial is
(set equal to the 4 here). Now we'll move the 4 back over and set the whole thing back equal to y:
From this it's apparent what the vertex is: (1, -4),
the axis of symmetry is x = 1, and
the y-intercept is found by setting the x's equal to 0 in the original equation and solving for y. So the y-intercept is (0, -3).
Your choice for the correct answer is the very last one there.