Help with 2 algebra problems?

Question

I am doing my Algebra II homework packet and I cannot get past these two problems I'm stuck on. Can someone help?

1. What number can be added to q[squared] - 30q to make a perfect square trinomial?

2. a) Consider the parabola with the equation y=4x[squared] - 12x + 7. Rewrite the equation in vertex form.

b) What is the vertex of this parabola?

Please help! I've gotten pretty far in the packet but I am really stuck on these two. 10 points to whoever answers the questions accurately first.

Answer 1

1. 15
2. a. y=(2x-3)^2 + 4
b. (3/2;4)

Answer 2

1). The general rule is: Take half the middle term
and square it. So -30/2 = -15; -15² = 225
Answer: 225. q² - 30q + 225 = (q-15)².

2). a). Factor out the 4 from the first 2 terms:
4(x²-3x) + 7
b). Make the term in parentheses a square, as in
problem 1:
4(x² -3x +9/4) + 7 -9.
4(x-3/2)² -2
That's your vertex form.
Part b). The vertex is therefore at (3/2, -2).
Hope that helps.
In the future, try to remember that rule for
completing the square!