Vertex form: y=a(x-h)^2+ky=a(x−h)2+k
(That's what vertex form is in the textbook here. Other forms are possible.)
Method 1
Your teacher probably wants you to complete the square. (It is an important technique, used for other things also.)
This method relies on the fact that: (x+-a)^2=x^2+- (2a)x+a^2(x±a)2=x2±(2a)x+a2
y=3x^2-18x+5=3(x^2-6x " --------" ) +5y=3x2−18x+5=3(x2−6x --------)+5 (Leave yourself some apce inside the parentheses)
If the stuff in parentheses is going to be a perfect square (x-a)^2(x−a)2, then it must have 2a=62a=6 which tells me that a=(1/2)(6)=3a=(12)(6)=3.
So the stuff inside parentheses is not a perfect square becuase it is missing a^2a2 which we would like to be 99.
Now, (within limits) we are in charge here! We will add 99 inside the parentheses to make it a perfect square
color(red) ("PROBLEM")PROBLEM 3(x^2-6x+9)3(x2−6x+9) is 3(x-3)^23(x−3)2, but it is not equal to what we started with.
color(green)("Solution")Solution We will add and subtract the 99 and regroup.
It looks like this:
y=3x^2-18x+5=3(x^2-6x " --------" ) +5y=3x2−18x+5=3(x2−6x --------)+5
=3(x^2-6x+9-9)+5=3(x2−6x+9−9)+5
Now regroup, keeping the perfect square were we want it to be. (We're in charge (within limits).)
y=3(x^2-6x+9)-3(9)+5y=3(x2−6x+9)−3(9)+5
(Convince yourself that this really is equal to what we started with. Do the algebra to simplify it.)
Now we'll write it this way:
y=3(x-3)^2-27+5y=3(x−3)2−27+5
y=3(x-3)^2-22y=3(x−3)2−22
And there's our answer.
(If your class uses a different vertex for, it's probably:
(y-k)=a(x-h)^2(y−k)=a(x−h)2 so the answer would be (y+22)=3(x-3)^2(y+22)=3(x−3)2)
Method 2
Depends on your knowing the vertex formula (which was 'discovered' by completing the square).
y=ax^2+bx+cy=ax2+bx+c has vertex with xx-coordinate (-b)/(2a)−b2a
(Why> Well, start with y=ax^2+bx+cy=ax2+bx+c and complete the square to get y=a(x+b/(2a))^2+"I'll leave it to you to find this"y=a(x+b2a)2+I'll leave it to you to find this
So y=3x^2-18x+5y=3x2−18x+5 has vertex at (-(-18))/(2(3))=3−(−18)2(3)=3
And when x=3x=3,
we have y=3(3)^2-18(3)+5=3(9)-6(9)+5=-3(9)+5=-27+5=-22y=3(3)2−18(3)+5=3(9)−6(9)+5=−3(9)+5=−27+5=−22
(I now, I do arithmetic kinda weird.)
So a=3a=3, h=3h=3 and k=-22k=−22
y=3(x-3)^2-22y=3(x−3)2−22
