Question 32675This question is from textbook
: This is my question:express each of the following in the form y = a(x - h)^2 + k.
Hence state the coordinate of the turning point and sketch the graph in each case:
a)x^2 - 2x + 3
b)x^2 - 8x +12
c)- x^2 + 4x + 1
d)x^2 +4x + 1
Can you show me the solution of these exercise.
This question is from textbook
Answer by kietra(57)   (Show Source):
You can put this solution on YOUR website! This is problem requires a longer explanation so I will only answer part A and then you can apply these same principles to the remaining 3 parts.
You have the problem y=x^2 - 2x + 3 and you want to manipulate it to the form y = a(x - h)^2 + k.
y=x^2 - 2x + 3 First, a will always be the coefficient on x^2- in this case, it is 1. To find the (x-h)^2 part, you will need to "complete the square".
(x - ?)(x - ?) will give you (x^2 - 2x + __ ) + 3. Forget about the three while caculate what should go in the parenthesis to give you (x^2 - 2x + __ ) when you factor it. You come up with (x - 1)(x - 1). WHen you foil this, you get (x^2 -2x + 1) and now, to keep the equation balanced, you need to subtract one. (x-1)^2 + 3 - 1. Combine those, and you get y = 1(x-1)^2 + 2
y = x^2 - 2x + 3
y = 1x^2 - 2x + 3
y = 1(x^2 - 2x + ?) + 3
y = 1(x - ?)(x - ?) + 3
y = 1(x^2 - 2x + 1) + 3 -1
y = 1(x - 1)(x - 1) + 3 -1
y = 1(x-1)^2 + 2
Now, we have it in the form y = a(x - h)^2 + k.
The VERTEX of the equation is at the point (h,k). Notice that the h is subtracted from the x so watch the sign carefully. In this case, the vertex would be (1,2). If a is positive, the vertex is a minimum and the parabola points upward (positive=happy face) and if a is negative, the vertex is a maximum (negative=frowny face). In this case, a=1 and is positive so the vertex is a minimum and the parabola points upward like a "U". To find the x interecepts, you substitute x=0 and solve for y.
This should be enough information for a basic sketch. Good luck with the remaining problems.
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